Quantum Magnetism, Spin Waves, and Light

Silvia Viola Kusminskiy

Max Planck Institute for the Science of Light Staudtstraße 2, 91058 Erlangen, Germany Abstract

Both magnetic materials and light have always played a predominant role in information technologies, and continue to do so as we move into the realm of quantum technologies. In this course we review the basics of magnetism and quantum mechanics, before going into more advanced subjects. Magnetism is intrinsically quantum mechanical in nature, and magnetic ordering can only be explained by use of quantum theory. We will go over the interactions and the resulting Hamiltonian that governs magnetic phenomena, and discuss its ele- mentary excitations, denominated magnons. After that we will study magneto- optical effects and derive the classical Faraday effect. We will then move on to the quantization of the electric field and the basics of optical cavities. This will allow us to understand a topic of current research denominated Cavity Optomagnonics. These notes were written as the accompanying material to the course I taught in the Summer Semester 2018 at the Friedrich-Alexander Universität in Erlan- gen. The course is intended for Master or advanced Bachelor students. Ba- sic knowledge of quantum mechanics, electromagnetism, and solid state at the Bachelor level is assumed. Each section is followed by a couple of simple exer- cises which should serve as to "fill in the blanks" of what has been derived, plus specific references to bibliography, and a couple of check-points for the main concepts developed. The figures are pictures of the blackboard taken during the lecture.

1 Contents

1 Electromagnetism 3 1.1 Basic magnetostatics ...... 3 1.2 Magnetic moment ...... 5 1.3 Orbital angular momentum ...... 7 1.4 Spin angular momentum ...... 8 1.5 Magnetic moment in a magnetic ﬁeld ...... 9 1.6 Magnetization ...... 13 1.7 Magnetostatic Maxwell equations in matter ...... 14 1.8 Demagnetizing ﬁelds ...... 16

2 Atomic Origins of Magnetism 20 2.1 Basics of quantum mechanics ...... 20 2.2 Orbital angular momentum in quantum mechanics ...... 21 2.3 Hydrogen atom ...... 22 2.4 Addition of angular momentum and magnetic moment ...... 23 2.5 Generalization to many electrons ...... 26

3 Magnetism in Solids 28 3.1 The Curie-Weiss law ...... 29 3.2 Exchange interaction ...... 32 3.3 Hydrogen molecule ...... 34 3.4 Heisenberg, Ising, and XY models ...... 37 3.5 Mean ﬁeld theory ...... 39 3.6 Ground state of the ferromagnetic Heisenberg Hamiltonian . . . 41 3.7 Ground state of the antiferromagnetic Heisenberg Hamiltonian . 43 3.8 Ground state of the classical Heisenberg model ...... 45 3.9 Dipole-dipole interactions ...... 46

4 Interlude: Problem Set 47

5 Spin Waves and Magnons 49 5.1 Excitations of the Heisenberg ferromagnet ...... 49 5.2 Equation of motion approach ...... 54 5.3 Holstein-Primakoﬀ transformation ...... 56

2 CONTENTS 3

5.4 Spin-wave approximation ...... 58 5.5 Magnon-magnon interactions ...... 60

6 Magneto-Optical Eﬀects 64 6.1 Electromagnetic energy and zero-loss condition ...... 64 6.2 Permittivity tensor and magnetization ...... 68 6.3 Faraday eﬀect ...... 69 6.4 Magneto-optical energy ...... 71

7 Modern Topics: Cavity Optomagnonics 73 7.1 Quantization of the electromagnetic ﬁeld ...... 73 7.2 Optical cavities as open quantum systems ...... 77 7.3 The optomagnonic Hamiltonian ...... 81 7.4 Coupled equations of motion and fast cavity limit ...... 83 Chapter 1

Electromagnetism

1.1 Basic magnetostatics

The history of magnetism is ancient, for example, the magnetic compass was invented in China more that 2000 years ago. The fact that magnetism is intrinsically connected to moving electric charges (and not "magnetic charges") however was not discovered until much later. In the year 1819, Oersted experimentally demonstrated that a current-carrying wire had an effect on the orientation of a magnetic compass needle placed in its proximity. In the following few years, Ampere realized that a small current loop generates a magnetic field which is equivalent to that of a small magnet, and speculated that all magnetic fields are caused by charges in motion. In what follows we will concentrate on magnetostatics, that is, magnetic field that are constant on time. The condition for that is a steady-state current, in which both the charge density ρ and the current density j = I/As (As cross-sectional area) are independent of time ∂ρ = 0 ∂t ∂j = 0 . ∂t From the continuity equation ∂ρ ∇ · j + = 0 ∂t we therefore obtain ∇ · j = 0 . In these notes we will call magnetic induction to B and magnetic field to H.1 In free space, these two fields are related by

B = µ0H (1.1.1)

1Some authors call instead B the magnetic ﬁeld and H the auxiliary ﬁeld.

4 CHAPTER 1. ELECTROMAGNETISM 5

−7 −2 being µ0 = 4π × 10 NA the permeability of free space. We will use the SI units system throughout these notes, therefore B is measured in Teslas (T = V.s.m−2) and H in Amperes per meter (A.m−1). The magnetic induction at point r due to a current loop can be calculated using the Biot-Savart law µ I r dB = 0 ds × (1.1.2) 4πr2 r where ds points in the direction of the current I, see Fig.1.1.1. Equivalent to Biot-Savart is Ampere’s law, which reads

B · ds = µ0I (1.1.3) ˛C where I is the current enclosed by the closed loop C, see Fig.1.1.1. Ampere’s law is general, but it is useful to calculate magnetic fields only in cases of high symmetry, for example an infinite straight wire. Using Stoke’s theorem, we can put Ampere’s law in differential form

∇ × B = µ0j . (1.1.4) Amperes law together with the absence of magnetic monopoles condition

∇ · B = 0 (1.1.5) constitute the Maxwell equations for magnetostatics.

Figure 1.1.1: The magnetic induction generated by a current I can be calculated using Biot-Savart’s law, see Eq.1.1.2. Ampere’s law (see Eq.1.1.3) is always valid, but useful to calculate the B ﬁelds only for cases of particular symmetry, e.g., an inﬁnite straight wire. CHAPTER 1. ELECTROMAGNETISM 6

Check points • What is the magnetostatic condition? • Write the magnetostatic Maxwell equations

1.2 Magnetic moment

The magnetic moment of a current loop is defined as m = IAnˆ , (1.2.1) where A is the area enclosed by the loop and nˆ is the normal to the surface, with its direction defined from the circulating current by the right-hand rule, see Fig.1.2.1. m defines a magnetic dipole in the limit of A → 0 but finite moment.

Figure 1.2.1: Magnetic dipole: the magnetic ﬁeld induced by a small current loop is equivalent to that of a small magnet.

Using Eq.1.1.2 we can calculate the magnetic induction generated by a small current loop of radius R µ I ds0 × ∆r B(r) = 0 (1.2.2) 4π ˆ ∆r3 µ I 1 = − 0 ds0 × ∇ 4π ˆ ∆r CHAPTER 1. ELECTROMAGNETISM 7

Figure 1.2.2: Magnetic induction due to a small circular current loop: we use Biot- Savart to calculate the B field. From "far away" it is the field of a magnetic dipole. with ∆r = r − r0 (see Fig.1.2.2). Therefore we can define B(r) = ∇ × A(r) . In the far field limit (∆r R ), µ ˆr A(r) = 0 m × (1.2.3) 4π r2 µ 3(m · r)r − r2m B(r) = 0 (1.2.4) 4π r5 which is the magnetic induction generated by a magnetic dipole. More generally, for an arbitrary current density distribution j(r0) one can define 1 m = d3r0 [r0 × j(r0)] (1.2.5) 2 ˆ and Eq.1.2.3 is the lowest non-vanishing term in a multipole expansion of the vector potential (in the Coulomb gauge, ∇ · A = 0) µ j(r0) A(r) = 0 d3r0 . (1.2.6) 4π ˆ |r − r0| The energy of a magnetic dipole in a magnetic field is given by

EZ = −m · B (1.2.7) and therefore is minimized for m k B. This is called the Zeeman Energy. CHAPTER 1. ELECTROMAGNETISM 8

1. Exercise: derive Eqs.1.2.3 and 1.2.4 (tip: use the "chain rule" and a multipole expansion). 2. Exercise: show that Eq.1.2.1 follows from 1.2.5 (tip: 1-D Delta- function distributions have units of 1/length). Recommended reading for Sections 1.1 and 1.2: • Ref. [6] (Jackson), Chapter 5 • Ref. [4] (Griﬃths), Chapter 5

Check points • How do you show the equivalence between the magnetic ﬁeld of a small current loop and that of a small magnet? A conceptual explanation suf- ﬁces.

1.3 Orbital angular momentum

The magnetic moment m can be related to angular momentum. For that, we consider the limit of one electron e (with negative charge −e) orbiting around a fixed nucleus, see Fig.1.3.1. Note that here we get the first indication that magnetism is a purely quantum effect: stable orbits like that are not allowed classically, and we need quantum mechanics to justify the stability of atoms. The average current due to this single electron is e eω I = − = − (1.3.1) T 2π where T is one period of revolution. The electron also possesses orbital angular momentum L = mer × v. Measured from the center of the orbit,

2 L = meR ~ω (1.3.2) and using Eq. 1.2.1 we obtain e m = − L . (1.3.3) 2me Therefore we have linked the magnetic moment of a moving charge to its orbital angular momentum. The coeﬃcient of proportionality is called the gyromagnetic ratio e γL = − , (1.3.4) 2me which is negative due to the negative charge of the electron. Hence in this case the magnetic moment and angular momentum are antiparallel. In solids, electrons are the primary source of magnetism due to their small mass compared 3 to that of the nucleus. Since mp ≈ 10 me , the gyromagnetic ratio for the nucleus is strongly suppressed with respect to the electronic one. CHAPTER 1. ELECTROMAGNETISM 9

Figure 1.3.1: Semiclassical picture of the orbital angular momentum of an electron.

Check points • What is the gyromagnetic ratio?

1.4 Spin angular momentum

Although we performed a classical calculation, the result obtained for the gyromagnetic ratio in Eq.1.3.4 is consistent with the quantum mechanical result. We know however that the electron posses an intrinsic angular momentum, that is, the spin S. The total angular momentum of the electron is therefore given by J = L + S . (1.4.1)

The spin has no classical analog and the coeﬃcient of proportionality γS between magnetic moment and spin mS = γSS (1.4.2) needs to be calculated quantum mechanically via the Dirac equation. The results is e γ ≈ − = 2γ , (1.4.3) S m L where the approximate symbol indicates that there are relativistic corrections (also contained in the Dirac equation!) to this expression. The γS value agrees with experimental observations. The total magnetic moment of the electron is therefore given by

mTOT ≈ γL (L + 2S) (1.4.4) CHAPTER 1. ELECTROMAGNETISM 10 and hence is not simply proportional to the total angular momentum! To understand the relation between mTOT and J, given by the Landé factor, we need to resort to quantum mechanics and the operator representation of angular momentum. We will do that in the next chapter, where we discuss the atomic origins of magnetism. Recommended reading: If you are interested in how γS is derived from the Dirac equation, check sections 2.2 and 2.3 in Ref. [11]"Quantum Theory of Magnetism" by Wolfgang Nolting and Anupuru Ramakanth.

Check points • What is the relation between the magnetic moment of an electron and the angular momentum operators?

1.5 Magnetic moment in a magnetic ﬁeld

A magnetic moment in a magnetic ﬁeld experiences a torque

T = m × B . (1.5.1)

Therefore, the classical equation of motion for the magnetic dipole (considering only the orbital angular momentum) is

dL = m × B = γ L × B . (1.5.2) dt L Using the geometry depicted in Fig.1.5.2 we obtain

L˙ = L sin θϕ˙eφ

γLL × B = |γL|LB sin θeφ .

Hence the magnetic moment precesses around B at a frequency

ωL =ϕ ˙ = |γLB| (1.5.3) which is denominated the Larmor frequency. Therefore the angular momentum will precess around the B field at a fixed angle θ and with constant angular frequency ωL. This is consistent with the energy expression defined in Eq.1.2.7. The work per unit time performed by the torque is given by the usual expression for the power dW = T· ~ω (1.5.4) dt where the angular velocity vector is perpendicular to the plane of rotation and its direction is given by the right hand rule. We see therefore that there is no power transfer in the Larmor precession, since ωLzˆ ·T = 0. There is however an energy cost if we want to change the angle θ of precession, since the resultant CHAPTER 1. ELECTROMAGNETISM 11

Figure 1.5.1: A magnetic moment in a B ﬁeld experiences a torque T = m × B. Remember: L and m point in opposite directions.

˙ angular velocity θeϕ is collinear with the torque. Using (for simplicity we deﬁned now θ as the angle between m and B, see Fig.1.5.3).

T = −mB sin θeϕ ˙ ~ωθ = −θeϕ (1.5.5) we ﬁnd dθ T θ˙ = −mB sin θ (1.5.6) dt and therefore the work exerted to rotate up to an angle θ is (up to a constant)

W = − mB sin θdθ = m · B = −E (1.5.7) ˆ Z .

We can therefore take the Zeeman energy EZ as the potential energy associated with the necessary work required to rotate the dipole m with respect to an external B ﬁeld. We will see equations of motion in the form of Eq.1.5.2 reappearing throughout this course, even as we treat the angular momenta as quantum operators. The reason is that, even though the total magnetic moment mTOT is not proportional to the total angular momentum J (see Eq.1.4.4), their quantum mechanical expectation values are proportional to each other through the Landé factor g. We will see this more formally when we start dealing with the quantum mechanical representation of the angular momenta. For now, we assume that mTOT and J are related by

mTOT · J = gγLJ · J (1.5.8) CHAPTER 1. ELECTROMAGNETISM 12

(a)

(b)

Figure 1.5.2: Coordinates used for solving the equation of motion. a) Side view, in the plane formed by L and B, and b)View from above, in a plane perpendicular to B. CHAPTER 1. ELECTROMAGNETISM 13

Figure 1.5.3: Coordinate system used to obtain Eq.1.5.7. CHAPTER 1. ELECTROMAGNETISM 14 from which we can obtain a classical expression for g, by replacing Eqs.1.4.4 and 1.4.1 into 1.5.8 and noting that 1 L · S = J 2 − L2 − S2 2 from J 2 = (L + S)2. One obtains

3 S2 − L2 g = + , (1.5.9) cl 2 2J 2 where the superscript indicates this is a classical approximation for g, which coincides with the quantum mechanical result in the limit J 2, S2, and L2 large. Recommended reading: Sections 1.2 and 1.3 from Ref. [12]"Spin Waves: Theory and Applications" by Daniel D. Stancil and Anil Prabhakar.

Check points • Write the equation of motion for an angular momentum in the presence of a magnetic ﬁeld

• What is the dynamics of an angular momentum in the presence of a magnetic ﬁeld? • What is the Larmor frequency?

1.6 Magnetization

Inside a material, the magnetic induction B indicates the response of the material to the applied magnetic ﬁeld H. Both vector ﬁelds are related through the magnetization in the sample

B = µ0 (H + M) , (1.6.1) where the magnetization is defined as the average magnetic moment per unit volume, hmi M(r) = V , (1.6.2) V and where the average indicates that we average over all atomic magnetic moments in a small volume V around position r. In this way, a smooth vectorial function of position is obtained. From Eq.1.6.1, we see the magnetization has the same units as the magnetic field H (A.m−1). In Eq.1.6.1, both B and H indicate the fields inside the material, so H contains the demagnetizing fields. We will see more on demagnetization fields in the next section. The response to the magnetic field of the magnetization and field induction are characterized by the magnetic susceptibility χ and the permeability µ, CHAPTER 1. ELECTROMAGNETISM 15 respectively M χ = (1.6.3) H B µ = , (1.6.4) H where we have written the simplest expressions for the case in which all fields are collinear, static (that is, independent of time) and homogeneous in space (q = ω = 0). In general however the response functions are tensorial quantities, P e.g. Mi = j χijHj, and depend on frequency ω and momentum q. Note that from Eq.1.6.1 we obtain µ µr = = 1 + χ µ0 again in the simple collinear case. µr is the relative permeability, is dimensionless and equals to unity in free space. The quantities defined in Eqs.1.6.3 and1.6.4 are still allowed to depend on temperature T and magnetic field H. We will now consider qualitatively the dependence on H. For linear materials, χ and µ are independent of H.A linear material with negative constant susceptibility is diamagnetic, whereas a positive susceptibility indicates either paramagnetism (no magnetic order) or antiferromagnetism (magnetic order with magnetic moments anti-aligned and zero total magnetization). In these cases, the magnetization is finite only in the presence of a magnetic field. On the other hand, if χ and µ depend on H, the relations Eqs.1.6.3 and1.6.4 are non linear. This is the case for magnetically ordered states with net magnetization, namely ferromagnets (magnetic moments aligned and pointing in the same direction) and ferrimagnets (magnetic moments anti-aligned but of different magnitude, so there is a net magnetization). In these materials, the magnetization increases non-linearly with the applied field and saturates when all magnetic momenta are aligned. When decreas- ing the magnetic field, there is a remanent, finite magnetization at zero field. This process is called hysteresis and it is used to magnetize materials. As we learned in the previous section, the magnetic momentum, and hence the magnetic characteristics of a material, are related to the total angular momentum of the electrons, and therefore on the atomic structure.

Check points • What is the relation between magnetic moment and magnetization?

1.7 Magnetostatic Maxwell equations in matter

To calculate the magnetic dipole moment m from Eq.1.2.5we have to know the microscopic current density. In general however we are not interested in microscopic, fast ﬂuctuations. We already saw an example in which we considered the average current I generated by one orbiting electron, to obtain semiclassically CHAPTER 1. ELECTROMAGNETISM 16

the gyromagnetic ratio γL in Sec.1.3. We have also deﬁned the magnetization M as a macroscopic quantity which entails the average density of the microscopic m. In a material, in general we have access to the magnetization, which is due to bound microscopic currents, and to the macroscopic current density due to free charges, which we will denominate jF. This motivates deﬁning a macroscopic vector potential A in terms of these two macroscopic quantities, and not the microscopic currents as in Eq.1.2.6

µ j (r0) M(r0) × (r − r0) A(r) = 0 d3r0 F + . (1.7.1) 4π ˆ |r − r0| |r − r0|3 Note that this is simply re-writing Eq.1.2.6, separating the bound- and free- currents contributions. The bound-current contribution, the second term in Eq.1.7.1, is written in terms of the magnetization and is equivalent to an averaged Eq.1.2.3. Eq.1.7.1 allows us to deﬁne an eﬀective current density associated with the magnetization, by noting that

0 0 3 0 M(r ) × (r − r ) 3 0 0 0 1 d r 0 3 = d r M(r ) × ∇ 0 (1.7.2) ˆV |r − r | ˆV |r − r | 0 0 3 0 0 0 1 M(r ) × da = d r ∇ × M(r ) 0 + 0 . ˆV |r − r | ˛S |r − r | (1.7.3)

We can therefore deﬁne an eﬀective bound volume current density

jB = ∇ × M (1.7.4) and an eﬀective bound surface current density

KB = M × nˆ (1.7.5) where the surface element is deﬁned as da = daˆn. In the bulk, for a well behaved magnetization function, the surface integral vanishes and we obtain

µ j (r0) j (r0) A(r) = 0 d3r0 F + B . (1.7.6) 4π ˆ |r − r0| |r − r0|

The surface current KB enters usually through boundary conditions at inter- faces. If we now go back to Ampere’s Eq.1.1.4 and separate the total current density into free and bound contributions j = jF + jB, we obtain

∇ × B = µ0 (jF + ∇ × M) (1.7.7) which defines the magnetic field H 1 H = B − M (1.7.8) µ0 CHAPTER 1. ELECTROMAGNETISM 17 such that ∇ × H = jF . (1.7.9) Therefore the magnetic field H takes into account in an average way the bound currents, and has as it’s only source the free currents. Eq.1.7.9 is equivalent to Eq.1.1.4, just re-written in a more convenient form for macroscopic magnetostatics in matter. Note that H is, on the contrary to B, not divergence-free:

∇ · H = −∇ · M . (1.7.10)

The magnetostatic Maxwell equations in matter therefore read

∇ · B = 0

∇ × H = jF (1.7.11) and have to be complemented by the constitutive equation B = µH in linear media or B = F (H) in non-linear (e.g., ferromagnetic) media, where F is a characteristic function of the material. We ﬁnish this section by stating the magnetostatic boundary conditions at an interface between two diﬀerent media 1 and 2

(B2 − B1) · nˆ = 0 (1.7.12)

nˆ × (H2 − H1) = KF (1.7.13) where KF is a free surface current density (usually 0). 1. Exercise: Prove Eqs.1.7.2, 1.7.3.

Check points • What is the meaning of the magnetic ﬁeld H? • What are the magnetostatic Maxwell equations in matter?

1.8 Demagnetizing ﬁelds

A crucial difference between magnetic and electric fields is the lack of free magnetic charges, or monopoles.2 They are however a useful mathematical con- struction in some cases, for example, to calculate the so called demagnetization fields. In finite systems, we can consider the magnetization as dropping to zero abruptly at the boundary of the material, giving rise to an accumulated ”magnetic charge density” at the surface which acts as an extra source of magnetic fields inside of the material. These fields in general oppose to an applied magnetic field and are therefore dubbed demagnetizing fields. A surface magnetic

2Magnetic monopoles, if they exist, have evaded experimental detection so far. They can however emerge as effective quasiparticles in condensed matter systems, and have been detected in materials which behave magnetically as a "spin ice". CHAPTER 1. ELECTROMAGNETISM 18 charge density is energetically costly, and we will see that for finite magnetic systems at the microscale it can determine the spatial dependence of the magnetically ordered ground state, giving rise to magnetic textures. If we consider the special case of no free currents, jF = 0, Eqs.1.7.11 imply we can define a magnetic scalar potential φM such that

H = −∇φM and using Eq.1.7.10 we obtain a Poisson equation

2 ∇ φM = −∇ · M with solution

0 ˆ0 0 1 3 0 ∇ · M(r ) 1 0 n · M(r ) φM(r) = − d r 0 + da 0 . (1.8.1) 4π ˆV |r − r | 4π ˛S |r − r | Analogous to the case of the vector potential in Eq.1.7.3, this allows us to deﬁne an eﬀective magnetic-charge density

ρM = −∇ · M (1.8.2) and an eﬀective magnetic surface-charge density

σM = M · ˆn . (1.8.3)

We see that ρM can only be finite for a non-homogeneous magnetization M(r), whereas finite σM indicates a discontinuity of M at the chosen surface S. 1. Exercise: Uniformly magnetized sphere For a ferromagnet at saturation, the magnetization can be considered as given, so we can in principle calculate the resulting magnetic field for a given geometry using Eq.1.8.1. We consider here as an example the case of an uniformly magnetized sphere as depicted in Fig.1.8.1. a) Choosing ˆz in the direction of M we can write M = M0zˆ. Calculate ρM and σM and write the Poisson equation for φM . b) Show that the scalar potential inside of the sphere is 1 φin = M z (1.8.4) M 3 0 and find the magnetic field Hin and magnetic induction Bin inside of the sphere. c) The magnetic field Hin inside of the sphere opposes the magnetization and it is therefore called a demagnetizing field. The proportionality coefficient between Hin and M is called the demagnetizing factor N. What is the value of N in this case? Demagnetization factors are geometry dependent, and can moreover be defined only in very special cases with simple geometries. Besides the sphere, one can define demagnetization factors for an infinite plane, an infinite cylinder, and a spheroid. CHAPTER 1. ELECTROMAGNETISM 19

Figure 1.8.1: Coordinate system for the uniformly magnetized sphere problem. CHAPTER 1. ELECTROMAGNETISM 20

d) Let’s assume now the sphere is placed in an external magnetic ﬁeld H0. Using linearity, write the solution for Hin and Bin in this case. e) Let’s now consider the case that the sphere is not permanently magnetized, but we now the material has a permeability µ. From the constitutive equation

Bin = µHin (1.8.5) obtain the magnetization as a function of the external magnetic ﬁeld Mµ(H0), where the notation Mµ implies that in this case we consider the magnetization not as given, but it depends on the permeability of the material. Show that Mµ(0) = 0, and therefore the obtained expression is not valid for materials with permanent magnetization. Recommended reading: Sections 5.8 to 5.11 from Jackson’s book "Classical Electrodynamics", Ref. [6].

Check points • Which is the origin of the demagnetization factors? Chapter 2

Atomic Origins of Magnetism

2.1 Basics of quantum mechanics

We ﬁrst review some basic concepts of quantum mechanics. In quantum mechanics, we describe a particle of mass m in a potential V by a wavefunction ψ(r, t) which satisﬁes the Schrödinger equation

∂ψ(r, t) 2 i = − ~ ∇2ψ(r, t) + V ψ(r, t) . (2.1.1) ~ ∂t 2m The probability of ﬁnding the particle at a time t in a volume element d3r around position r is given by |ψ(r, t)|2d3r . If the potential V is independent of time, ψ(r, t) = ψ(r)f(t) and ψ(r) is an eigenfunction of the time-independent Schrödinger equation

2 − ~ ∇2ψ(r) + V (r) ψ(r) = Eψ(r) (2.1.2) 2m with energy E. Equivalently, we can write the eigenvalue equation for the Hamil- tonian in Dirac notation Hˆ |ψi = E |ψi (2.1.3) where the quantum state of the particle is represented by the ket |ψi, the respective wavefunction is ψ(r) = hr|ψi, and the Hamiltonian operator is

pˆ2 Hˆ = + V (ˆr). (2.1.4) 2m

In the position representation, ˆp → −i~∇ so that, for example, hr|pˆ|ψi = −i~∇ψ(r) . In general, for any operator Aˆ we can write the eigenvalue equation Aˆ |ψαi = α |ψαi where |ψαi is an eigenstate with eigenvalue α. For a hermitian operator, Aˆ† = Aˆ, α is real and the eigenstates form a basis of the Hilbert space where the operator acts. This is called an observable. The expectation value for Aˆ if the

21 CHAPTER 2. ATOMIC ORIGINS OF MAGNETISM 22

system is in the eigenstate |ψαi then is simply hψα| Aˆ |ψαi = α. If we consider a second operator Bˆ acting on the same Hilbert space, it is only possible to ﬁnd a common basis of eigenstates of Aˆ and Bˆ if and only if the two operators commute: [A,ˆ Bˆ] = AˆBˆ − BˆAˆ = 0. In this case, the two operators can be measured simultaneously to (in principle) arbitrary precision. If the operators do not commute, then we run into the Heisenberg uncertainty principle. The most well known example is that of the momentum and position operators, which satisfy [ˆx, pˆ] = i~. How precise we measure one of the operators will determine the precision up to which we can know the value of the other: ∆x∆p ≥ ~/2.

2.2 Orbital angular momentum in quantum mechanics

The orbital angular momentum in quantum mechanics is inherited from its classical expression, Lˆ = ˆr × pˆ. In the position representation is is given by

Lˆ = −i~r × ∇ . (2.2.1) From this expression it is easy to verify that the diﬀerent components of Lˆ do not commute with each other. Instead, one obtains ˆ ˆ ˆ [Li, Lj] = i~ijkLk , (2.2.2) where ijk is the Levi-Civita tensor and the Einstein convention for the implicit sum or repeated indices has been used. Therefore, it is not possible to measure simultaneously with arbitrary precision all components of the angular momentum. Let’s assume we choose to measure Lˆz . In this case, the Heisenberg uncertainity principle reads

∆L ∆L ≥ ~|hL i| . (2.2.3) x y 2 z It is however possible to ﬁnd a common basis for Lˆ2 and one of the angular 2 momentum components, since [Lˆ , Lˆi] = 0. Typically, Lˆz is taken and the respective eigenvalues are labelled by l and m. These are called the quantum numbers. The eigenstates satisfy ˆ2 2 L |ψlml i = ~ l(l + 1) |ψlml i ˆ Lz |ψlml i = ~m |ψlmli . (2.2.4)

For the orbital angular momentum, l is an integer and −l ≤ ml ≤ l. These conditions can be depicted pictorically as in Fig.2.2.1. The angular momentum p vector has a magnitude ~ l(l + 1) and its projection on the z-axis is quantized and takes one of the possible values ~ml. The maximum value of Lz is ~l, p instead of ~ l(l + 1) as one would expect classically. We recover the classical expectation in the limit l 1 . The Lx and Ly components do not have a deﬁnite value and are represented as a precession of L around the z-axis. CHAPTER 2. ATOMIC ORIGINS OF MAGNETISM 23

Figure 2.2.1: Pictorial representation of the spatial quantization of angular momentum. In this example, l = 3. L precesses around the z axis and has a deﬁnite projection on it, that can take one of the allowed values −3 ≤ ml ≤ 3.

Check points • Explain graphically the properties of an angular momentum operator and how it diﬀers from a classical angular momentum.

2.3 Hydrogen atom

For a problem with rotational symmetry the angular momentum is conserved: [H,ˆ Lˆ] = 0. Hence we can ﬁnd a basis of eigenstates that are also energy eigenstates, such that Hˆ |ψnlmi = En |ψnlmi . (2.3.1) The quantum numbers are denominated as principal, azimuthal, and magnetic respectively for n, l, and m. If we ignore spin, we have all the tools to solve the energy levels and orbitals for the hydrogen atom, in which the electron is subject to the Coulomb potential

e2 VC (r) = − (2.3.2) 4πε0r due to the nucleus. Due to the spherical symmetry of the problem, is is convenient to write the wavefunction in spherical coordinates. From Eq.2.1.2 it can be shown that ψnlm(r, θ, φ) = Rnl(r)Ylm(θ, φ) CHAPTER 2. ATOMIC ORIGINS OF MAGNETISM 24

2 The principal number n = 1, 2, 3... gives the quantization of energy En ∝ −1/n . The Rnl(r) are associated Laguerre functions and determine the radial profile of the probability distribution for the electron. Ylm(θ, φ) are spherical har- monics, which can also be written in terms of the associated Legendre functions, m imφ Ylm(θ, φ) = Pl (cos θ)e . For m = 0, Yl0(θ) = Pl(cos θ) are simply the Legen- dre polynomials. The azimuthal number l = 0, 1, ..., n−1 labels the usual s,√ p, d, ... orbitals. The s orbitals are spherically symmetric, since Y00(θ, φ) = 1/ 4π. The higher the azimuthal number, the higher the probability to find the electron further away from the nucleus, whereas n gives the number of nodes of the wavefunction in the radial direction. That the azimuthal quantum number l is quantized was demonstrated experimentally in what we now know as the Zeeman effect. For that, we come back to the relation between the orbital angular momentum and the magnetic moment. We can write p mL = µB l(l + 1)

mL · z = −µBml (2.3.3) where µB = ~γL is the Bohr magneton and the expressions correspond to expectation values. If the atom is placed in an external magnetic ﬁeld, there will be an extra contribution to the energy due to the Zeeman term, see Eq.1.2.7. We can take the z-axis to coincide with the magnetic ﬁeld, hence

EZ = µBmlB (2.3.4) and we say that the degeneracy of the l level, originally 2l + 1, is split for all l > 0. This splitting can be measured in the absorption spectrum of atoms with total spin angular momentum equal to zero. To add the eﬀect of the spin degree of freedom, we have however ﬁrst to understand how to combine angular momentum operators in quantum mechanics.

Check points • What are the quantum numbers for the Hydrogen atom and what do they tell us?

2.4 Addition of angular momentum and magnetic moment

The orbital angular momentum is the generator of rotations in position space. In general however, we can deﬁne an angular momentum simply by its algebra, determined by the commutation relation 2.2.2. The spin angular momentum CHAPTER 2. ATOMIC ORIGINS OF MAGNETISM 25 generates rotations in spin space and satisﬁes ˆ ˆ ˆ [Si, Sj] = i~ijkSk 2 2 Sˆ |si = ~ s(s + 1) |si ˆ Sz |si = ~ms |si . In contrast to the orbital angular momentum, the spin quantum number s is not constrained to be an integer, and can take also half-integer values. Fermions (e.g., the electron) have half-integer values of spin and bosons integer values. For electrons, s = 1/2 and ms = ±1/2 . The spin operator commutes with the orbital angular momentum

[Sˆ, Lˆ] = 0 since they act on diﬀerent state spaces. One would be therefore tempted to n 2 2 o choose Sˆ , Lˆ , Lˆz, Sˆz as a set of commuting observables. Spin and orbital angular momentum however interact via the spin-orbit interaction, a relativistic correction to the Hamiltonian 2.1.4 which reads

2 ˆ ~ 1 ∂V ˆ ˆ HSO = 2 2 S · L . (2.4.1) 2mec r ∂r For an atomic system, V is the Coulomb potential. This correction is usually small, but it increases with the atomic number. Since Sˆ and Lˆ are coupled by Eq.2.4.1, they are not conserved and they do not commute separately with the Hamiltonian. The total angular momentum Jˆ = Sˆ + Lˆ however is conserved. n 2 2 2 o We choose therefore Sˆ , Lˆ , Jˆ , Jˆz as a set of commuting observables, and s, l , j and mj are our quantum numbers. The change of basis is achieved via the Clebsch-Gordan coeﬃcients X |sljmji = |sljmji hsmslml|sljmji . (2.4.2)

msml

We now have the task of relating the magnetic dipolar moment mTOT to the total angular momentum operator Jˆ. From Eq.1.4.4 we know that mTOT is not collinear with Jˆ, but it is proportional to Lˆ + 2Sˆ. We note however that the projection mˆ TOT · Jˆ is well deﬁned, since

2 2 2 [Lˆ · Jˆ, Jˆ ] = [Lˆ · Jˆ, Lˆ ] = [Lˆ · Jˆ, Sˆ ] = [Lˆ · Jˆ, Jˆz] = 0 2 2 2 [Sˆ · Jˆ, Jˆ ] = [Sˆ · Jˆ, Lˆ ] = [Sˆ · Jˆ, Sˆ ] = [Sˆ · Jˆ, Jˆz] = 0 . (2.4.3)

The magnetic moment therefore precesses around Jˆ, as depicted in Fig.2.4.1. The Wigner-Eckart theorem allows us to relate ˆmTOT with Jˆ in terms of expectation values by noting that

ˆ ˆ 0 ˆ 0 γLhsljmj|L + 2S|sljmji = γLg(slj)hsljmj|J|sljmji . CHAPTER 2. ATOMIC ORIGINS OF MAGNETISM 26

Figure 2.4.1: Pictorial depiction of the precession of Lˆ + 2Sˆ = ˆmTOT/γL around Jˆ.

Therefore in the (2j + 1) degenerate subspace with ﬁxed quantum numbers (s, l, j) we can think of ˆmTOT as being proportional to Jˆ. In the literature, this is sometimes denoted as " ˆmTOT = γLgJˆ " in an abuse of notation. The value of g(slj) we can ﬁnd by projecting mˆ TOT · Jˆ

2 hsljmj| Lˆ + 2Sˆ · Jˆ|sljmji = γLg(slj)hsljmj|Jˆ |sljmji and using Eqs.2.4.3 we obtain the Landé factor 3 s(s + 1) − l(l + 1) g(s, l, j) = + . (2.4.4) 2 2j(j + 1) We see that this expression coincides with the classical one obtained in Eq.1.5.9 in the limit of s, l, j 1. Going back to the Zeeman splitting, we see that in general the Zeeman correction to the atomic energy in the presence of a magnetic ﬁeld is given by

EZ = µBg(s, l, j)mjB (2.4.5) and therefore the Zeeman splitting depends on all the atomic orbital numbers, and it is not simply µBB. This is denominated the anomalous Zeeman effect since at the time of the experiments the spin was still not known, and it was not possible to explain the effect. Note however that the normal Zeeman effect can be observed only for atoms with zero total spin, and therefore the anomalous one is much more common. 1. Exercise: Prove Eqs.2.4.3 2. Exercise: Derive Eq.2.4.4. CHAPTER 2. ATOMIC ORIGINS OF MAGNETISM 27

Check points • How do you relate the magnetic moment of an electron to its total angular momentum? Why? • What is the Landé factor?

2.5 Generalization to many electrons

To generalize the previous concepts beyond the hydrogen atom / single electron problem is impossible to do in an exact manner, since the problem turs into a many-body problem: the many electrons interact not only with the nucleus, but among themselves. We can however make analytical progress by doing some reasonable approximations. The first one is called the Hartree approximation, in which consider that each electron moves in an effective central potential Veff (r) generated by the nucleus plus all the other electrons. The other electrons are said to screen the potential of the nucleus, since their charge is opposite. The second approximation concerns the spin-orbit coupling Eq.2.4.1. For all except the heaviest atoms, the spin-orbit interaction is weak and can be treated within perturbation theory. In this case, we can first neglect this interaction, an consider that Lˆ i and Sî for each electron i are independent. Hence, we can calculate the total orbital and spin angular momentum simply by summing them ˆ P ˆ ˆ P ˆ separately: LTOT = i Liand STOT = i Si , and then proceed to calculate the total angular momentum JˆTOT = Lˆ TOT + SˆTOT and the corresponding ˆ2 P magnetic moment. STOT has eigenvalues ~S(S + 1) with S = i ms,i and, ˆ 2 P respectively, LTOT has eigenvalues ~L(L + 1) with L = i ml,i. The allowed values of JˆTOT are given by the angular momentum summation rules: J : |L − S|, |L − S| + 1, ...L + S. This approximation is denominated the Russell- Sanders coupling and it gives rise to the well known Hund’s rules. To calculate JˆTOT we need first a prescription to obtain SˆTOT and Lˆ TOT. A closed shell means that we have occupied all 2(2l + 1) levels in it, where the factor of 2 comes for the spin s = ±1. Therefore both the total orbital and spin angular momentum of the shell are zero, and hence also the total angular momentum Jˆ in the shell. The total angular momentum of the atom, and therefore its magnetic properties, will be determined by the last, partially unoccupied shell. If all shells are closed (that is, full), then JˆTOT = 0 and the atom is diamagnetic. The Hund rules tell us how to distribute our n ≤ 2(2l + 1) "left over" electrons in the last shell.The total spin SˆTOT we obtain by applying the Pauli principle: since electrons are fermions, their wavefunction is antisymmetric and two electrons can not have the same quantum numbers. Each orbital characterized by l can then be occupied by only two electrons, one with spin up, and one with spin down. The first Hund rule tells us to maximize S, since this will tend to put one electron with in each orbital until half filling, (2l + 1), and then continue with spin down. This minimizes Coulomb repulsion by putting electrons as far apart as possible. The second Hund rule tells us to maximize the orbital angular momentum, once the spin is maximized. This CHAPTER 2. ATOMIC ORIGINS OF MAGNETISM 28 also minimizes Coulomb repulsion, by making the electrons orbit as far apart as possible. The third Hund rule sounds more mysterious: if the shell is more than half filled (n ≥ 2l+1) then J = L+S, and for a less than half filled shell (n ≤ 2l+ 1), J = |L−S| . This is actually due to the spin-orbit interaction λLˆ TOT ·SˆTOT, where λ’s sign changes between these two configurations. Therefore to minimize spin-orbit coupling requires SˆTOT and Lˆ TOT parallel or antiparallel, depending on the filling. To finish this section, we point out that for the heavier elements the Russell- Sanders coupling prescription is not valid anymore, due to the strong spin-orbit coupling. For these elements, a different prescription, denominated jj coupling, is used. There Jî for each electron is first calculated, and then the total JˆTOT. As we mentioned, if all shells are closed the total angular momentum is zero and the atom is diamagnetic. Diamagnetism can be understood by the Faraday law: as a magnetic field is turned on, it induces a change in the orbital motion of the electrons which opposes the change in magnetic flux. Diamagnetism is usually a weak effect and it is overshadowed by paramagnetism in atoms with partially unfilled shells. Once the atoms are ordered in a lattice and form a solid, it can happen that magnetic order develops as the temperature is lowered. This will depend on the electronic interactions, as we will see in the next chapter. Recommended reading for Secs.2.4 and 2.5: Chapter 31 from Ashcroft and Mermin’s book "Solid State Physics", Ref. [1].

Check points • What is the Russell-Sanders coupling scheme? Chapter 3

Magnetism in Solids

In the previous chapter we showed how to calculate the magnetic moment of an atom. We saw that the problem is already quite involved even for a single atom if we go beyond a hydrogen-like one. When atoms come together to form a solid, to treat the magnetic problem atom per atom is not only impossible but also wrong, since we have to take into account the binding between the atoms that form the solid, and what matters is the collective behavior of the material. In a solid, the orbitals of the constituent atoms overlap to form bands instead of discrete energy levels. Depending on the character of the orbitals involved in the magnetic response of a material, we can divide the problem into two big subsets: metals and insulators. In metals, the orbitals are extended and have a good amount of overlap, so the electrons are delocalized and free to move around the solid. In insulators, the orbitals are narrow and we talk about localized magnetic moments. Of course, this is an oversimpliﬁed view: there are systems in which both localized and delocalized electrons participate in magnetism, or in which magnetism and electric conduction are due to diﬀerent groups of orbitals. An example is that of heavy fermion systems, which can be modeled as a Kondo Lattice: a lattice of magnetic impurities embedded in a sea of free electrons. Itinerant magnetism (that one due to delocalized electrons) and mixed systems belong to what one calls strongly correlated systems, highly complex many-body problems in which electron-electron interactions have to be taken into account. In this course we will concentrate mostly on magnetic insulators. Moreover, to understand magnetic ordering we have to introduce electronic interactions and take into account the Pauli principle. Recommended reading for this chapter: besides what is indicated in the sub- chapters, Chapter 5 of the book "Quantum Theory of Magnetism" by Wolfgang Nolting and Anupuru Ramakanth, Ref. [11].

29 CHAPTER 3. MAGNETISM IN SOLIDS 30

3.1 The Curie-Weiss law

We consider ﬁrst a system of localized, N identical non interacting magnetic moments, and calculate their collective paramagnetic response.1 We can obtain the magnetization from the Helmoltz free energy

F = −kBT ln Z (3.1.1) N ∂F M = − (3.1.2) V ∂B where kB is the Boltzmann constant, T is the temperature, and Z the canonical partition function for one magnetic moment

X − En Z = e kBT . (3.1.3) n If we consider a magnetic moment with total momentum J, we have 2J + 1 possible Jz values and J 1 X − gµBJz B Z = e kBT . (3.1.4)

Jz =−J The sum can be performed since it is a geometric one. Putting all together, one ﬁnds N µBgJB M = gµBJBJ ( ) (3.1.5) V kBT where 2J + 1 2J + 1 1 1 B (x) = coth( x) − coth( x) J 2J 2J 2J 2J is the Brillouin function. For µBB kBT this function goes to 1 and the magnetization saturates: all momenta are aligned with the B-ﬁeld. For large temperatures instead, µBB kBT , one obtains the inverse-temperature dependence of the magnetization known as the Curie law, characterized by the susceptibility 2 M N µ0 (gµB) J(J + 1) χc = = . (3.1.6) H V 3 kBT

Experimentally, it is found that this result is good for describing insulating crystals containing rare-earth ions (e.g. Yb, Er), whereas for transition metal ions in an insulating solid agreement is found only if one takes J = S. This is due to an effect denominated angular momentum quenching, where effectively L = 0 . This quenching is due to crystal fields: since now the atoms are located in a crystalline environment, rotational symmetry is broken and each

1As we saw in the last chapter, all atoms present a diamagnetic behavior, but since it is very weak compared to the paramagnetic response, only atoms with closed shells (e.g. noble gases) prsent an overall diamagnetic response. An exception is of course superconducting materials, which can have a perfect diamagnetic response. This is however a collective, macroscopic response due to the superconducting currents which oppose the change in magnetic flux. CHAPTER 3. MAGNETISM IN SOLIDS 31 atom is located in electric fields due to the other atoms in the crystal. In the case of rare-earth ions, the magnetic moments come from f-shells, which are located deep inside the atom and therefore better isolated form crystal fields. For transition metals instead, the magnetic moments come from the outermost d-shells, which are exposed to the symmetry breaking fields. These however are spatial dependent fields and therefore do not affect the spin degree of freedom. As a consequence, the orbital angular momentum is not conserved and precesses around the crystal fields, averaging to zero. The Curie law is followed by all materials with net magnetic momentum for large enough temperature. For low temperatures however, in certain materials magnetic order develops and there is a deviation from the Curie law in the susceptibility. Weiss postulated the existence of molecular fields, which are proportional to the magnetization in the material and are responsible for the magnetic ordering. The total magnetic field acting on a magnetic moment within this picture is therefore Htot = Hext + λM and from the Curie susceptibility M M C χc = = = (3.1.7) Htot Hext + λM T with N µ (gµ )2 J(J + 1) C = 0 B (3.1.8) V 3 kB we obtain CH M = ext (3.1.9) T − λC and therefore the susceptibility in the external field follows the Curie-Weiss law C χ = . (3.1.10) W T − λC We see that for large temperatures this susceptibility follows the Curie inverse law, but as the temperature approaches a critical temperature TC = λC, χW diverges indicating a phase transition to the magnetically ordered phase. A magnetically ordered phase implies M 6= 0 for an external field Bext = 0. From Eq.3.1.5 we see that, if we do not consider the molecular fields postulated by Weiss, M(0) = 0. Let’s now consider Bext = 0 but the existence of a molecular field BW = µ0λM. Inserting this field in Eq.3.1.5 we obtain an implicit equation for M:

µBµ0gJλM M = M0BJ ( ) , (3.1.11) kBT where we have deﬁned the saturation magnetization N M = gµ J (3.1.12) 0 V B since BJ (x → ∞) = 1 (see Fig.3.1.1). Eq.3.1.11 has still a solution M = 0. CHAPTER 3. MAGNETISM IN SOLIDS 32

Figure 3.1.1: Brillouin function BJ

However, if d(M B ) 0 J ≥ 1 (3.1.13) dM M=0 we see that a second solution to the transcendental equation is possible. From J + 1 B (x → 0) ≈ x J 3J we obtain Msp Msp = λC TC and therefore TC = λC, in agreement with Eq.3.1.10. The subscript sp indicates this is a spontaneous magnetization, not induced by an external magnetic field. For T > TC , Eq.3.1.13 is not fulfilled and M = 0 is the only solution. This can be seen graphically in Fig.3.1.2. At the time, the origin of these postulated molecular fields was not known. Naively, one could expect the dipole-dipole interaction between the magnetic moments to be the origin of the magnetic ordering. This energy scale is however too small to explain magnetism at room temperature. The potential energy of one magnetic dipole m2 in the magnetic field created by another dipole m1, Vpot = −m2 · B1(r) is

2 µ0 3(m1 · r)r − r m1 Vpot = −m2 · 5 . (3.1.14) 4πε0 r A quick estimate corresponds to taking the distance between the dipoles as the interatomic distance, and the dipolar moments simply as Bohr magnetons. CHAPTER 3. MAGNETISM IN SOLIDS 33

Figure 3.1.2: Possibility of spontaneous magnetic order according to the condition Eq.3.1.13. (Note that in the main text we denominated Msp what in the ﬁgure is Mord).

Equating this energy to kBTC results in a critical temperature for magnetic ordering of TC ≈ 1K . Therefore dipolar interactions could be responsible magnetic ordering only below this temperature, which is very low. Instead, a rough estimate of the repulsive Coulomb energy between two electrons gives

2 Uc e 5 = 2 ≈ 10 K kB kB4πε0a which is very large! In the following section we will see that magnetic ordering is due to a combination of electrostatic energy and a very quantum eﬀect: the Pauli principle. Recommended reading: Chapter 31 from Ashcroft and Mermin’s book "Solid State Physics", Ref. [1]. 1. Exercise: Prove Eq.3.1.5.

2. Exercise: estimate TC from the dipole-dipole interaction.

Check points • What is the Curie law and when is it valid? • What is the Curie-Weiss law? CHAPTER 3. MAGNETISM IN SOLIDS 34

3.2 Exchange interaction

Let’s consider ﬁrst the case of two electrons subject to a Hamiltonian

2 2 p1 p2 H = + + V (r1, r2) . (3.2.1) 2me 2me This Hamiltonian is independent of spin, however the Pauli exclusion principle imposes a spatial symmetry on the wavefunction ψ(r1, r2) solution of Hψ(r1, r2) = Eψ(r1, r2). Since two electrons with the same quantum numbers are not allowed at the same place, the total wavefunction ψ(r1, r2; s1, s2) must be antisymmetric with respect to exchange of both spin and space:

ψ(r1, r2; s1, s2) = −ψ(r2, r1; s2, s1) .

If there is no spin-orbit coupling in our Hamiltonian, we can separate the wavefunction ψ(r1, r2; s1, s2) = ψ(r1, r2)χ(s1, s2) and therefore we see that the antisymmetry implies a correlation between the spin and orbital parts of the wavefunction. This is a constraint of the problem and solutions that do not fulﬁll this constraint are inﬁnite in energy. For two electrons, we have the total S = 0, 1 and correspondently Sz = 0, −1, 0, 1. We can write the corresponding states in explicit symmetric and antisymmetric linear combinations, the antisymmetric singlet state 1 |0; 0i√ (| ↑↓i − | ↓↑i) (3.2.2) 2 and the symmetric triplet state with S = 1, Sz = −1, 0, 1 :

|1; 1i = | ↑↑i (3.2.3) 1 |1; 0i = √ (| ↑↓i + | ↓↑i) 2 |1; −1i = | ↓↓i where the states are labeled as |S; Szi. Since the Hamiltonian does not depend on spin, we can work simply with the spatial component of the wavefunction, but imposing the right symmetry. The spatial part of the wavefunction corresponding to the singlet conﬁguration, ψsinglet(r1, r2) will therefore be symmetric in space coordinates, whereas ψtriplet(r1, r2) has to be antisymmetric

ψsinglet(r1, r2) = ψsinglet(r2, r1)

ψtriplet(r2, r1) = −ψtriplet(r2, r1) , (3.2.4) and we can write the full state as CHAPTER 3. MAGNETISM IN SOLIDS 35

|Ψsingleti = |ψsingleti|0; 0i

|Ψtriplet(Sz)i = |ψtripleti|1; Szi (3.2.5)

The Schrödinger equation therefore reads

H|ψsingleti = Es|ψsingleti

H|ψtriplet(Sz)i = Et|ψtripleti and if Es 6= Et, the ground state is spin-dependent even though H is not. If that is the case, since spatial and spin sectors are correlated, we search for a Hamiltonian operating in spin space H˜ , coupling ˆs1 and ˆs2 and such that it is equivalent to H:

H˜ |0; 0i = Es|0; 0i (3.2.6)

H˜ |1; Szi = Et|1; Szi .

The Hamiltonian that does the trick is 1 1 H˜ = (Es + 3Et) − (Es − Et)(ˆs1 · ˆs2) (3.2.7) 4 ~2 since ( ˆs · ˆs 1 3 − 3 S = 0 1 2 = S(S + 1) − = 4 . (3.2.8) 2 2 4 1 ~ 4 S = 1 We have therefore constructed a Hamiltonian acting on spin space which is in principle equivalent to the interacting Hamiltonian of Eq.3.2.1, which gives an eﬀective interaction between the spins. This is called the molecular Heisenberg model and can be written as

H˜ = J0 − J12ˆs1 · ˆs2 with 1 J12 = (Es − Et) . (3.2.9) ~2 If J12 > 0, this interaction favors a ferromagnetic alignment of the spins, consistent with the fact that the singlet energy is higher than the triplet one. We now show an example in which Es 6= Et and calculate explicitly J12. We will see that part of J12 has no classical analog and comes from interchanging particles 1 and 2, since in quantum mechanics they are indistinguishable.

1. Exercise: Prove Eq.3.2.8 and show that H˜ given in Eq.3.2.7 satisﬁes 3.2.6. CHAPTER 3. MAGNETISM IN SOLIDS 36

Check points • Why can you write a Hamiltonian in spin space which is equivalent to the Hamiltonian deﬁned in position space? • How do you impose the equivalence for the two-electrons system?

3.3 Hydrogen molecule

We consider now two hydrogen atoms that are brought close together to form a hydrogen molecule. We consider the nuclei, a and b, as ﬁxed at positions Ra and Rb, whereas the electrons, at positions r1 and r2 are subject to the nuclei Coulomb potential plus the repulsive Coulomb interaction between them . This corresponds to taking the nuclei mass ma, mb → ∞ and it is an example of the denominated Born Oppenheimer approximation. The Hamiltonian for each, separate hydrogen atom a and b are given by

2 2 ~∇1 e 1 Ha = − − 2me 4πε0 |Ra − r1| 2 2 ~∇2 e 1 Hb = − − 2me 4πε0 |Rb − r2| and we know the respective eigenfunctions, φa,b with eigenenergies Ea,b. If the distance between the two atoms |Ra − Rb| → ∞, these are the exact solutions. If however the atoms are brought close together to form a molecule, there will be an interaction term e2 1 e2 1 e2 1 e2 1 HI = − − − − 4πε0 |Ra − Rb| 4πε0 |Rb − r1| 4πε0 |Ra − r2| 4πε0 |r2 − r1| and the total Hamiltonian is given by

Htot = Ha + Hb + HI .

Even though this is a quite simple system, this problem cannot be solved exactly and we have to resort to approximations. We treat HI as a perturbation and use the atomic wavefunctions φa,b as a basis for a variational solution of the full wavefunction. This implies we are assuming the electrons are quite localized at their respective atom. This approach is denominated the Heitler-London method. If we consider the unperturbed Hamiltonian

H0 = Ha + Hb the eigenfunctions will be simply linear combinations of the product of the original orbitals, φaφb, with eigenenergy Ea + Eb since the two systems do not interact with each other. To preserve the indistinguishability of the particles, we cannot simply write a solution as φa(r1)φb(r2), since the probability density CHAPTER 3. MAGNETISM IN SOLIDS 37

2 2 |φa(r1)| |φb(r2)| is not invariant under exchanging r1 ←→ r2. The symmetric and antisymmetric combinations fulﬁll however the indistinguishability condition 1 ψs(r1, r2) = √ (φa(r1)φb(r2) + φb(r1)φa(r2)) 2 1 ψt(r1, r2) = √ (φa(r1)φb(r2) − φb(r1)φa(r2)) (3.3.1) 2 where with the notation ψs, ψt we have anticipated that the symmetric (antisymmetric) solution in space corresponds to the singlet (triplet) solution in spin space, with a full wavefunction as given in Eq.3.2.5. Note that both ψs, ψt are eigenfunctions of H0 with eigenvalue Ea + Eb, and therefore without the interaction HI , Et = Es = Ea + Eb and the soulutions are 4-fold degenerate. We now calculate Et = Es perturbatively in the presence of the interaction HI and using ψs, ψt as variational wavefunctions

hψs/t|Htot|ψs/ti Es/t = . (3.3.2) hψs/t|ψs/ti

We are interested in the ground state solutions, so that φa,b are solutions of each hydrogen atom with Ea,b = E0. The variational principle tells us that the energies calculated by Eq.3.3.2 are always greater or equal that the true groundstate.2 Let’s ﬁrst analyze the simple overlap

hψ |ψ i = d3r d3r ψ∗ (r , r )ψ (r , r ) . (3.3.3) s/t s/t ˆ 1 2 s/t 1 2 s/t 1 2

If the two atoms are inﬁnitely apart |Ra − Rb| → ∞, then the orbitals corresponding to diﬀerent atoms have zero overlap, hφa|φbi = 0, and

hψ |ψ i = d3r d3r |φ (r )|2|φ (r )|2 = 1 . (3.3.4) s/t s/t 0 ˆ 1 2 a 1 b 2

When the atoms are brought close together, their orbitals will overlap: hφa|φbi= 6 0. We do not calculate this overlap explicitly, but simply note that it will be ﬁnite and denote it by O

O2 = d3r d3r φ∗(r )φ (r )φ (r )φ∗(r ) (3.3.5) ˆ 1 2 a 1 b 1 a 2 b 2 and hence 2 hψs/t|ψs/ti = 1 ± O . (3.3.6)

2 Note that in general the variational procedure would be to write ψs/t(r1, r2) = c1φa(r1)φb(r2) ± c2φb(√r1)φa(r2) and find the coefficients c1,2 by minimizing Eq.3.3.2. One then finds c1 = c2 = 1/ 2. In Eq.3.3.1 we used our knowledge of the symmetry of the problem plus the normalization of the wavefunctions φa,b to write the result immediately. CHAPTER 3. MAGNETISM IN SOLIDS 38

We now turn to the numerator in Eq.3.3.2. We already know that the non- interacting contribution is

hψs/t|H0|ψs/ti = 2E0 . (3.3.7) hψs/t|ψs/ti

The correction to this non-interacting energy is given by hψs/t|HI |ψs/ti, which we see contains two kind of terms. One of them is simply the Coulomb electrostatic interaction between the two atoms, assuming that they are close enough to interact but electron 1(2) still "belongs" to atom a(b)

K = d3r d3r φ∗(r )φ (r )H φ (r )φ∗(r ) . (3.3.8) ˆ 1 2 a 1 b 2 I a 1 b 2 The remaining term has no classical analog, and it measures the Coulomb energy cost upon exchanging the two electrons

X = d3r d3r φ∗(r )φ (r )H φ (r )φ∗(r ) . (3.3.9) ˆ 1 2 a 1 b 1 I a 2 b 2 and it is called the exchange integral, or exchange interaction. Putting all together we obtain K ± X E = 2E + (3.3.10) s/t 0 1 ± O2 where the + (−) corresponds to the singlet (triplet) solution ψs (ψt ). In general O 1 and we can replace the denominator by 1. We have therefore shown that Es − Et 6= 0 and hence for this problem 2 J12 ≈ X, (3.3.11) ~2 which justiﬁes the name exchange parameter for J12.

Check points • What is the Heitler-Lindon model? • What is the exchange interaction and why does it not have a classical analog?

3.4 Heisenberg, Ising, and XY models

In our variational solution for the Hydrogen molecule from the last section, double occupation is forbidden, so two electrons cannot be in the same atom at the same time. This indicates our treatment is valid for insulators, where electrons are quite localized, but in turn leads necessarily to small values of the exchange parameter, since J12 relies on the overlap of the single-atom orbitals. The above example therefore must be taken as a toy model which reveals the character of CHAPTER 3. MAGNETISM IN SOLIDS 39 the ferromagnetic interaction. In general, the exchange constant is generated by more complex interactions, e.g. superexchange where the ferromagnetic exchange interaction between two spins is mediated by an exchange interaction with an atom in between those with a net angular momentum. We further postulate that our model can be generalized to N multielectron atoms 1 X H = − J Sˆ · Sˆ (3.4.1) 2 ij i j ij where the exchange coeﬃcient Jij is taken as a parameter of the model, that has to be calculated for each particular material. The Hamiltonian in Eq.3.4.1 is the Heisenberg Hamiltonian. The factor of 1/2 accounts for the double-counting in the sum. We write Sˆ by convention, and we refer to the magnetic moment as "spins" in an abuse of language: in reality, unless the orbital angular momentum is quenched, the total angular momentum of the ions is meant. The spin operators follow the angular momentum algebra when located at the same site, and commute with operators at diﬀerent sites:

ˆα ˆ β ˆγ [Si , Sj ] = i~δijαβγ Si (3.4.2) where α, β, γ indicate the spatial components of the angular momentum x, y, z. These commutation relations make the quantum Heisenberg model, despite its simple appearance, quite a rich model and exactly solvable only for a few simple cases. The input of the model is the lattice from and dimensionality, and the exchange parameter Jij . Besides insulators, the Heisenberg Hamiltonian is a valid model for localized magnetic moments embedded in a metal. In that case, the exchange interaction is mediated by the conduction electrons, which gives rise to the RKKY interaction (Ruderman–Kittel–Kasuya–Yosida). The calculated exchange function Jij is an oscillating function of position, alternating between positive and negative values, and is longer ranged than in the insulating case. If there are no local magnetic moments but the system still present magnetic order, the conduction electrons are also responsible for the magnetic order. In this case the magnetism is denominated itinerant and it is described by a different Hamiltonian: the Hubbard Hamiltonian. This model takes into account the kinetic energy of the electrons, who can "jump" from lattice site to lattice site, and penalizes double occupation with a local Coulomb repulsion term. The Hubbard model takes an effective Heisenberg form in the particular case of a half-filled band and strong Coulomb interaction. The Heisenberg Hamiltonian is the "father" Hamiltonian of other well known models in magnetism. In a crystal, crystal fields can give rise to anisotropies in the exchange parameter Jij. If the anisotropy is along only one direction one can write X ˜ ˆx ˆ x ˆy ˆ y ˆz ˆ z H = − Jij Si Sj + Si Sj + ∆ Si Sj . (3.4.3) ij CHAPTER 3. MAGNETISM IN SOLIDS 40

If ∆ > 1, magnetic ordering occurs along the z axis, which is denominated the easy axis. For ∆ 1 the Hamiltonian turns eﬀectively into the Ising Model

X z z HIsing = − Jij Si Sj . (3.4.4) ij

Note that in this particular case, all operators in the Hamiltonian commute, and therefore the model is in this sense classical. If ∆ < 1 then we have an easy plane ordering. For ∆ 1 the system is eﬀectively two-dimensional and isotropic, which is termed the XY model

X ˆx ˆ x ˆy ˆ y HXY = − Jij Si Sj + Si Sj . (3.4.5) ij If the system is placed in an external magnetic ﬁeld, a Zeeman term is added to the Hamiltonian 1 X X H = − J Sˆ · Sˆ − gµ B · Sˆ 2 ij i j B i ij i 1 X = − J Sˆ · Sˆ − gµ BSˆz . (3.4.6) 2 ij i j B i ij

In Eq. (3.4.6) the spin operator are dimensionless and ~ has been absorbed in µB (correspondingly, in the commutation relation Eq. (3.4.2) ~ should be set to 1 if this convention is used). Note that the Zeeman energy is −m · B (see Eq. (1.2.7)) and tends to align the magnetic moment with the magnetic ﬁeld, and anti-align the angular momentum. Sometimes by convention the extra minus sign is not used, since from now onwards one always works with the angular momentum operators, and one takes gµB > 0. This corresponds simply to transform B → −B if we want to translate into the magnetization or magnetic moments, and does not aﬀect the results.

Check points • Write the Heisenberg Hamiltonian in a magnetic ﬁeld

3.5 Mean ﬁeld theory

Once we have a Hamiltonian that models our system, we want in principle (i) find the ground state, that is, the lowest energy eigenstate of the system, which is the only state populated at zero temperature, (ii) find the excitations on top of this ground state, which will determine the behavior of the system at T 6= 0, and (iii) study phase transitions, either at T = 0 (denominated a quantum phase transition) where by changing some other external parameter like the magnetic field, the ground state of the system changes abruptly, or at finite temperature where an order parameter of the system, given by a quantum-statistical average CHAPTER 3. MAGNETISM IN SOLIDS 41 of some relevant quantity to describe the system, goes to zero as a function of temperature or other external parameters, e.g. the magnetization M(B,T ). We go back now to the issue of magnetic ordering armed with the Heisenberg Hamil- tonian. As we pointed out above, this is a very rich and complicated model, and there are very few general statements that can be made about the three points above. We turn therefore to a well know approximation denominated mean field theory. This approximation is good only for long-range interactions and high dimensions, it is however widely used to get a first idea of, for example, what kind of phases and phase transitions our model can present. For the Heisenberg model, we will see that mean field theory will give as a microscopic justification of the Weiss molecular fields we introduced at the beginning of this chapter. We start by assuming that hSîi is finite. For example, for a ferromagnetic ground state, hSîi is uniform and such that N M = gµ hSˆ i , (3.5.1) V B i where N is the number of lattice sites and V the total volume. We write now Sî in the suggestive form Sî = hSîi + Sî − hSîi , (3.5.2) which corresponds to splitting the operator into its quantum statistical average value and the fluctuations with respect to this average. In mean field theory, these fluctuations are assumed to be small. The mean field Hamiltonian is obtained from the Heisenberg Hamiltonian by keeping only terms up to first order in the fluctuation, 1 X X X H = J hSˆ i · hSˆ i − J hSˆ i · Sˆ − gµ B · Sˆ , (3.5.3) MF 2 ij i j ij i j B i ij ij i where we already included an external magnetic field B. The first term in Eq. (3.5.3) is simply a constant shift in the energy. The second term corresponds to a spin a site i in the presence of a magnetic field generated by all other spins. We can therefore define an effective magnetic field 1 X B = B + J hSˆ i eff gµ ij i B i and our problem is reduced from an interacting problem, to that of non-interacting spins in the presence of a magnetic field. If we go back to Eq. (3.1.7), we see that now we can give a microscopic explanation to the Weiss molecular fields, which where assumed to be proportional to to the magnetization. In particular we find that we can write λ Beff = B + M µ0 with 1 V 1 λ = 2 J0 µ0 N (gµB) CHAPTER 3. MAGNETISM IN SOLIDS 42

P where we have used translational symmetry and deﬁned J0 = i Jij independent of j. This system therefore presents a transition to an ordered state as a function of temperature and magnetic ﬁeld as discussed for the Curie-Weiss law, but we now have a microscopic explanation for the phenomenological model.

Check points • What is the meaning of the mean ﬁeld theory? • How is it related to the Curie-Weiss law?

3.6 Ground state of the ferromagnetic Heisen- berg Hamiltonian

For the particular case in which Jij ≥ 0 ∀ i, j it is possible to find the ground state of the Heisenberg Hamiltonian without further specifications. This is called the ferromagnetic Heisenberg model since the ground state is ferromagnetically ordered, as we show in the following. We consider hence the Hamiltonian 1 X Hˆ = − J Sˆ · Sˆ , with J = J ≥ 0 . (3.6.1) 2 ij i j ij ji ij if the spins were classical vectors, the state of lowest energy would be that one with all N spins are aligned. Hence a natural candidate for the ground state of Hˆ is |0i = |S, Si1|S, Si2...|S, SiN (3.6.2) ˆ2 ˆz where Si |S, Sii = S(S + 1)|S, Sii and Si |S, Sii = S|S, Sii, that is all spins take their maximum projection of Sˆz (we consider all spins identical). The individual spin operators however do not commute with the Hamiltonian 3.6.1, h i Sî, Hˆ 6= 0 and therefore a product states of the form

|ψi = |S, m1i|S, m2i...|S, mN i (3.6.3)

ˆz (with Si |S, mii = mi|S, mii) span a basis of the Hilbert space, but are not necessarily eigenstates of Hˆ . The total spin operator SˆTOT however does commute ˆ with H h i SˆTOT, Hˆ = 0

ˆ ˆ2 ˆz and we can construct an eigenbasis for H, STOT, and STOT. The state 3.6.2 is a state with maximum SˆTOT. We will now prove that (i) 3.6.2 is an eigenstate of 3.6.1, and (ii) that there is no eigenstate with higher energy. CHAPTER 3. MAGNETISM IN SOLIDS 43

To follow with the proof it is convenient to recast Hˆ in term of ladder operators

ˆ± ˆx ˆy Si = Si ± iSi (3.6.4) ˆ± p Si |Si, mii = (Si ∓ mi)(Si + 1 ± mi)|Si, mi ± 1i , from which it is clear why they are call ladder operators: Sˆ+ (Sˆ−) increases (decreases) the projection of Sˆz by one unit. In terms of 3.6.4, the Hamiltonian 3.6.1 reads 1 X 1 Hˆ = − J Sˆ+Sˆ− + Sˆ−Sˆ+ + SˆzSˆz . (3.6.5) 2 ij 2 i j i j i j ij With this expression it is now straightforward to show that |0i is an eigenstate of Hˆ 1 X S 2 X Hˆ |0i = − J SˆzSˆz|0i = − J |0i 2 ij i j 2 ij ij ij ˆ+ since Si |0i = 0 ∀ i (remember that spin operators at diﬀerent sites commute, and Jii = 0). Therefore Hˆ |0i = E0|0i with S 2 X E = − J . 0 2 ij ij

We now need to prove that E0 is the minimum possible energy. For that we consider the expectation value of Hˆ with an arbitrary product state as in Eq. 3.6.3 1 X 1 X E0 = hψ|Hˆ |ψi = − J hψ|SˆzSˆz|ψi = − J m m 0 2 ij i j 2 ij i j ij ij where we have used that

We further note that if Jij ≥ 0 then X X Jijmimj ≤ JijSS ij ij and hence 0 E0 ≥ E0 , which proves that the fully polarized state |0i given in Eq. 3.6.2 is indeed the ground state. This ground state is however not unique, but it is (2STOT + 1) degenerate in spin space, with STOT = NS. This can be easily visualized in the CHAPTER 3. MAGNETISM IN SOLIDS 44

2-spins case, where for the triplet state, STOT = 1 and the state has 3 possible projections of Sˆz, see Eq. 3.2.3. Note that each of these states is moreover infinitely degenerate in position space, since we are able to choose the quantization axis freely. This is an example of what is called spontaneous symmetry breaking, which occurs when the ground state has a lower symmetry than the Hamiltonian. The ground state of Hˆ is a specific realization of the (2STOT + 1) possible ground states, and therefore has "picked" a preferred direction in spin space, which is not determined by the symmetry of Hˆ . We mention in passing that our results are valid in all its generality strictly for T = 0. For T > 0, the Mermin-Wagner theorem tells us that, in 1 and 2 dimensions and for short range interactions continuous symmetries cannot be spontaneously broken. If we add a magnetic field to Hˆ our total Hamiltonian is the one in Eq. 3.4.6. In this case, it is favorable for the system also to maximize the projection of Sˆz, and in this case |0i given in Eq. 3.6.2 is the only ground state, with energy

S 2 X E (B) = − J − gµ BNS. (3.6.6) 0 2 ij B ij In this case the symmetry is not spontaneously broken, since is is already broken at the Hamiltonian level by the applied magnetic ﬁeld. Recommended reading: Chapter 33 from Ashcroft and Mermin’s book "Solid State Physics", Ref. [1]. For those interested in knowing more about spontaneous symmetry breaking, Chapter 6 of the book "Interacting Electrons and Quantum Magnetism" by Assa Auerbach, Ref. [2].

Check points • What is the ground state of the ferromagnetic Heisenberg Hamiltonian? • How do you prove it is the ground state? • What happens in the presence of an external magnetic ﬁeld to the degeneracy of the state?

3.7 Ground state of the antiferromagnetic Heisen- berg Hamiltonian

Aside from the ferromagnetic case treated in the previous section, finding the ground state of the Heisenberg Hamiltonian is in general difficult and has to be studied case by case. The ground state will depend on the nature of the interactions (short or long range, sign, anisotropy), from the lattice structure, and from the dimensionality of the system. To illustrate this difficulty, we discuss here briefly the antiferromagnetic Heisenberg model on a bipartite lattice. In this case Jij ≤ 0 ∀ i, j and the lattice can be sub-divided in two sublattices A and B, such that Jij is finite only when i and j belong to two different sublattices. The simplest example is that of a square lattice with nearest neighbor CHAPTER 3. MAGNETISM IN SOLIDS 45

Figure 3.7.1: Antiferromagnetic Heisenberg model on a bipartite square lattice, the dotted lines indicate nearest-neighbor interactions. Example of a Néel state. interactions, where all nearest neighbors of A belong to the B sublattice, and vice-versa, see Fig. A guess of the ground state based on on the classical model is the so called Néel state: the two sublattices are fully polarized, but in opposite directions Y Y |0?iAF = |S, Sii |S, −Sij . i∈A j∈B It is easy to see however that this state is not an eigenstate of the Heisenberg Hamiltonian. Using the representation of Hˆ in terms of ladder operators, see Eq. 3.6.5, we see that

1 X X 1 Hˆ |0?i = − J Sˆ+Sˆ−|0?i + Sˆ−Sˆ+|0?i + SˆzSˆz|0?i . AF 2 ij 2 i j AF i j AF i j AF i∈A i∈B The ﬁrst term is simply zero, since the sublatticesA and B are fully polarized in "the right way" with respect to the operators. The last term is simply proportional to |0?iAF. However for the second term

ˆ− ˆ+ Si Sj |0?iAF ∝ |S, i...|S, −S + 1ij...|S, S − 1ii...|S, mN i which shows that |0?iAF cannot be an eigenstate. In general, one can prove that the true ground state |0iAF is non-degenerate and a singlet of total spin: SˆTOT|0iAF = 0. This is called the Marshall’s Theorem. Which kind of singlet is the actual ground state is however not determined. Recommended reading: Section 5.1 of the book "Interacting Electrons and Quantum Magnetism" by Assa Auerbach, Ref. [2]. CHAPTER 3. MAGNETISM IN SOLIDS 46

Check points • What is a Néel state? • Why it is not the ground state of the antiferromagnetic Heisenberg Hamil- tonian?

3.8 Ground state of the classical Heisenberg model

The classical Heisenberg model is obtained by replacing the spin operators Sî by simple vectors Si with fixed length |Si| = S. As we saw in previous sections, this kind of approximation is valid in the limit of large spin S. For the classical model it is straightforward to find the ground state for translational invariant systems, Jij = Jji = J(Ri − Rj) (3.8.1) where Ri indicates the points on a Bravais lattice. In Fourier space 1 X ik·Ri Si = √ Ske (3.8.2) N k 1 X −ik·Ri Sk = √ Sie N i and hence

1 0 X X ik·Ri ik ·Rj Hcl = − J(Ri − Rj) e Sk · Sk0 e 2N 0 RiRj kk

Deﬁning ∆R = Ri − Rj and using

X ik·Ri e = Nδk,0 (3.8.3)

Ri being δk,k0 the Kronecker Delta, we obtain

1 0 0 X X ik·Ri ik ·Ri −ik ·∆R Hcl = − J(∆R) e Sk · Sk0 e e 2N 0 Ri∆R kk

1 X X −ik0·∆R = − J(∆R) Nδ 0 S · S 0 e 2N k+k ,0 k k ∆R kk0 1 X X = − J(∆R) S · S eik·∆R 2 k −k ∆R k and deﬁning X J(k) = J(∆R)eik·∆R (3.8.4) i CHAPTER 3. MAGNETISM IN SOLIDS 47 we obtain 1 X H = − J(k)S · S (3.8.5) cl 2 k −k k Imposing the constraint S2 = S2, one can show that the minimum of the energy given by Eq. 3.8.5 is given by setting k = Q, where Q determines the global maximum of J(Q) [see D. H. Lyons and T. A. Kaplan, Ref. [8]]. One obtains the equilibrium conﬁguration

Si = S (cos(Q.Ri), sin(Q.Ri), 0) which is in general a planar helical state. Since Q is determined by the maximum of J(Q), the order is not necessarily commensurate with the lattice, unless this maximum coincides with a high symmetry point in the Brillouin zone. Special cases are Q = 0, where we recover ferromagnetic order, or Q taking a value at the edge of the Brillouin zone, in which case the order is antiferromagnetic. Applying a magnetic ﬁeld in the z direction tilts the magnetic order out of plane.

Check points • What is the classical Heisenberg model and how does one in general ﬁnd its ground state?

3.9 Dipole-dipole interactions

In the Heisenberg Hamiltonian we have not included the dipole-dipole interaction term 2 " ˆ ˆ # ˆ µ0 (µBg) X 1 3Si · (ri − rj) Sj · (ri − rj) ˆ ˆ Hd−d = − 3 2 − Si · Sj . 4π 2 |ri − rj| |ri − rj| i6=j (3.9.1) We argued this term is too small to justify magnetic ordering at the observed temperatures, and we found out that the exchange interaction is orders of magnitude larger. However the exchange interaction depends on the overlap of orbitals, and as such is generally short-ranged: it decays exponentially with distance, at least for insulators. From Eq. 3.9.1 we see that the dipole-dipole interaction decays instead algebraically, as 1/r3, and that means that it will be dominant at large distances. This gives rise, in large samples, to the formation of domains. In intermediate-size samples, usually micrometer sized, the com- petition between dipole-dipole interactions and exchange interactions leads to textured ground states. Being long-ranged, the dipolar interactions are sensitive to the boundaries of the sample, where the spins try to align with the boundary, so as to minimize the "magnetic monopoles" at the surface as introduced in Chapter 1. Dipole-dipole interaction terms can be included in the Heisenberg Hamiltonian in the long-wavelength limit as demagnetization ﬁelds. Chapter 4

Interlude: Problem Set

1) Larmor Diamagnetism

Consider a system of non-interacting electrons in a constant magnetic ﬁeld B0 = ∇×A. In the Coulomb gauge, ∇·A = 0 and the vector potential can be written as 1 A = B × r . (4.0.1) 2 0 The kinetic energy of the system is therefore

n 1 X 2 T = (p + eA) (4.0.2) 2m i e i=1 being pi = mevi − eA the canonical momentum. a) We can elevate these quantities to operators. Show that, in the Coulomb h i gauge, ˆpi, Aˆ i = 0.

b) Show that for B0 = (0, 0,B0),

n 1 e2B2 X T = T + µ L · B + 0 x2 + y2 0 B 0 8m i i ~ e i with n 1 X T = p2 . 0 2m i e i=1 c) We have not considered the spin of the electrons. Write the Hamiltonian Hˆ of the non-interacting system including spin. d) Give an expression for the magnetic moment operator ˆ mˆ = −∇B0 H,

48 CHAPTER 4. INTERLUDE: PROBLEM SET 49

where the gradient is taken with respect to the magnetic ﬁeld: ∇B0 = ( ∂ , ∂ , ∂ ). What is the expectation value hmˆ i in terms of macroscopic ∂B0x ∂B0y ∂B0x quantities? e) Argue that for closed shells, the magnetic susceptibility χ is negative. A negative susceptibility indicates that the system is diamagnetic. Give an estimate for χdiam and compare it to typical values of paramagnetic susceptibilities. Recommended reading: Section 3.2 of the book "Quantum Theory of Mag- netism" by Wolfgang Nolting and Anupuru Ramakanth, Ref. [11].

2) Thermodynamic functions

The first law of thermodynamics for a magnetic system can be written as dU = T dS − V MdB + µdN where U = hHˆ i is the internal energy, T is the temperature, V the volume, M the magnetization, B the magnetic field, µ the chemical potential, and N the number of particles. h...i indicates the statistical expectation value. The Helmholtz free energy is defined by F = U−TS and has as independent variables F (T,B,N), dF = −SdT − V MdB + µdN . In the following we take N constant, so that dN = 0. a) Consider ∆U = hHˆ (B) − Hˆ (0)i and T = 0. Show that dU = −V MdB b) Consider the canonical partition function

−βHˆ ZN = ZN (T,B) = Tr e with β = 1/kBT . The internal energy in terms of the partition function is given as 1 U = hHˆ i = Tr Heˆ −βHˆ ZN Show that 1 d(−βF ) = β hHˆ idT − h∇ Hˆ idB T B ∂ hHˆ i = k T 2 ln Z B ∂T N 1 ∂ h∇ Hˆ i = − ln Z B β ∂B N c) From the previous expressions, conclude that

F (T,B) = −kBT ln ZN (T,B)

d) Give an expression for M in terms of ZN . Recommended reading: Section 1.5 of the book "Quantum Theory of Mag- netism" by Wolfgang Nolting and Anupuru Ramakanth [11]. Chapter 5

Spin Waves and Magnons

5.1 Excitations of the Heisenberg ferromagnet

In the last chapter we dealt with the ground state of the Heisenberg model for a few solvable examples. The ground state is the only eigenstate that is occupied at strictly zero temperature. The excitations on top of this ground state will determine the behavior of the system at low temperatures. For the Heisenberg ferromagnet at T = 0 we found that all spins are aligned with and the magnetization is given by the saturation magnetization N M = gµ S. s B V At T 6= 0 some spins will "ﬂip" or, more generally, decrease by one unit, which in turn will decrease the magnetization from it saturation value. An obvious candidate for an excited state is therefore

50 CHAPTER 5. SPIN WAVES AND MAGNONS 51 not a good approximation to an excited eigenstate, since ﬂipping a spin in such a manner has a very high energy cost, of the order of the exchange interaction. For example, for nearest neighbor interaction with exchange constant J, ﬂipping a spin has an energy cost of ∆E ∼ zJS, where z is the coordination number (that is, z is the number of nearest neighbors, e.g. for a square lattice z = 4). We have already seen that the exchange constant is of the order of the critical temperature, quite a high energy for regular ferromagnets. One can prove that an actual eigenstate of 3.6.5 is given by

1 X |ki = √ e−iRi·k|ii , (5.1.3) N Ri where Ri are the lattice sites. We see therefore that an eigenstate is formed by distributing the "flipped" spin over all sites, and therefore it is a collective excitation of the system. This collective excitation has well defined momentum ~k (up to a reciprocal lattice vector) and energy ~ω(k) and we call it a quasiparticle. In particular for magnetic systems these quasiparticles are denominated magnons. In the following we will show that |ki is an eigenstate of 3.6.5 and will calculate its dispersion relation ~ω(k). In order to do this, we first perform a Fourier transform of the Hamiltonian 3.6.5 by defining the spin operators in momentum space (in analogy to the classical case, see Eqs. 3.8.4)

1 X Sˆα = √ e−ik·Ri Sˆα (5.1.4) k N i Ri 1 ˆα X ˆα ik·Ri Si = √ Sk e N k where α = x, y, z, ±. Note that

† ˆ+ ˆ− Sk = S−k † ˆz ˆz Sk = S−k and one can verify that the commutators now read h i Sˆ+ , Sˆ− = 2Sˆz (5.1.5) k1 k2 k1+k2 h i Sˆz , Sˆ± = ±Sˆ± . k1 k2 k1+k2

Using that Jij = Jji and that spin operators at diﬀerent sites commute, we re-write slightly the Heisenberg Hamiltonian 3.6.5 as

1 X h i Hˆ = − J Sˆ+Sˆ− + SˆzSˆz . (5.1.6) 2 ij i j i j ij CHAPTER 5. SPIN WAVES AND MAGNONS 52

From Eqs. 5.1.4 and using again Eq. 3.8.1 we obtain

1 0 0 ˆ X X ˆ+ ik·Ri ˆ− ik ·Rj ˆz ik·Ri ˆz ik ·Rj H = − J(Ri − Rj) Sk e Sk0 e + Ske Sk0 e 2N 0 RiRj kk gµ B X X z ik·Ri − √ B0 Sˆ e N k Ri k where we added an external magnetic ﬁeld. We use ∆R = Ri − Rj and Eq. 3.8.3 to write

1 0 0 √ ˆ X X X i(k+k )·Ri ˆ+ ˆ− ˆz ˆz −ik ·∆R X ˆz H = − J(∆R) e Sk Sk0 + SkSk0 e − gµB NB0 Skδk,0 2N 0 ∆R Ri kk k √ 1 X X + − z z −ik0·∆R z − J(∆R)N δ 0 Sˆ Sˆ + Sˆ Sˆ 0 e − gµ NB Sˆ 2N k+k ,0 k k0 k k B 0 0 ∆R kk0 1 X X √ − J(∆R) Sˆ+Sˆ− + SˆzSˆz eik·∆R − gµ NB Sˆz . 2 k −k k −k B 0 0 ∆R k

Using the deﬁnition for J(k) given in Eq. 3.8.4 we ﬁnally obtain an expression for the Heisenberg Hamiltonian in momentum space

1 X √ Hˆ = − J(k) Sˆ+Sˆ− + SˆzSˆz − gµ NB Sˆz . (5.1.7) 2 k −k k −k B 0 0 k

1 X 1 X − J(k)Sˆ+Sˆ− |0i = − J(k) Sˆ− Sˆ+ + 2Sˆz |0i 2 k −k 2 −k k 0 k k 1 X √ X = − J(k) 2Sˆz |0i = −S N J(k)|0i = 0 2 0 k k where the last equality stems from

X X X ik·∆R X J(k) = J(∆R)e = J(∆R)Nδ∆R,0 = NJ(∆R = 0) = 0 . k k ∆R ∆R CHAPTER 5. SPIN WAVES AND MAGNONS 53

For the second term in Eq. 5.1.7

1 X 1√ X − J(k)SˆzSˆz |0i = − NS J(k)Sˆzδ |0i 2 k −k 2 k −k,0 k k 1√ = − NSJ(k = 0)Sˆz|0i 2 0 1 = − NS2J(k = 0)|0i . 2 We now note that X X X X NJ(k = 0) = N J(∆R) = J(∆R) = J(Ri − Rj) = Jij

∆R Ri∆R RiRj ij and therefore 1 X S2 X − J(k)SˆzSˆz |0i = − J |0i . 2 k −k 2 ij k ij For the third term in Eq. 5.1.7 we obtain simply √ ˆz −gµB NB0S0 |0i = −gµBNB0S|0i .

Putting all together we obtain Hˆ |0i = E0|0i with

S 2 X E (B ) = − J − gµ B NS 0 0 2 ij B 0 ij which coincides with our result in Eq. 3.6.6 as expected. We now want to show that |ki given in Eq. 5.1.3 is a eigenstate of 5.1.7. Using 5.1.2 we can write |ki in terms of the ladder operators in momentum space, 1 |ki = √ Sˆ−|0i , (5.1.8) 2S k ˆ− hence it is enough to show that Sk |0i is an eigenstate. Writing

h ˆ ˆ−i ˆ− we see that that amounts to showing that H, Sk |0i ∝ Sk |0i. Using similar manipulations as above, and taking into account that J(k) = J(−k), one obtains

h ˆ ˆ−i ˆ− H, Sk |0i = [gµBB0 − S (J(k) − J(k = 0))] Sk |0i (5.1.9) and therefore |ki is an eigenstate of 5.1.7 with eigenvalue

E(k) = E0(B0) + gµBB0 − S (J(k) − J(k = 0)) . (5.1.10) CHAPTER 5. SPIN WAVES AND MAGNONS 54

The energy on top of the ground state is the excitation energy

~ω(k) = gµBB0 − S [J(k) − J(k = 0)] (5.1.11) which is simply the energy of one magnon — the quasiparticle energy is always defined with respect to the ground state energy, in field theory it is common to call the ground state of a system the "vacuum". Eq. 5.1.11 is also called the dispersion relation, since it gives the dependence of the energy with the wave vector. Note that for B0 = 0, ~ω(0) = 0 and therefore any infinitesimal temperature will cause excitations with k = 0. This "zero mode" is an example of a Goldstone mode, which is always present when there is spontaneous symmetry breaking in the system. In general, Goldstone modes are massless bosons quasiparticles which appear in systems where there is a spontaneously broken continuous symmetry. If we compare with the ground state energy E0(B0), in particular the Zeeman term, we see that the magnetic moment of the system has been modified by one unit. We can therefore conclude that a magnon has spin 1, and therefore it is ˆz a bosonic quasiparticle. The expectation value of a local spin operator Si with respect to the one-magnon state |ki can be shown to be 1 1 hk|Sˆz|ki = S − = h0|Sˆz|0i − ∀ i, k (5.1.12) i N i N which shows that the spin reduction is indeed of one unit and it is distributed uniformly over all sites Ri. Semiclassically, one can picture each spin in the lattice precessing around the z axis with a projection of ~(S − 1/N). However, except for k = 0, the spins do not precess in phase but instead they have a phase difference of e−ik·(Ri−Rj ), forming a spin wave. This semiclassical picture can be better understood if we look at the equations of motion for the spins, which we do in the following section. We finishing this section by stating that our intuition fails again if we want to construct two-magnon states. The obvious choice 0 ˆ− ˆ− |k, k i ∝ Sk Sk0 |0i (5.1.13) is actually not an eigenstate of the Heisenberg Hamiltonian. This is due to magnon-magnon interactions present in the Hamiltonian, which are not taken into account in a simple product state as 5.1.13. The same of course holds for multiple-magnon states. Therefore, when having multiple magnons excited in a system, they can interact and magnon states will decay due to magnon- magnon interactions. These excited states therefore have a certain lifetime. Besides magnon-magnon interactions, scattering with impurities or phonons in a material will determine the lifetime of magnon states. Recommended reading: Section 7.2.2 of the book "Quantum Theory of Mag- netism" by Wolfgang Nolting and Anupuru Ramakanth, Ref. [11]. 1. Exercise: Prove Eq. 5.1.9. 2. Exercise: Prove Eq. 5.1.12. CHAPTER 5. SPIN WAVES AND MAGNONS 55

Check points • What is a magnon? • How do you obtain the energy dispersion of a magnon?

5.2 Equation of motion approach

In the Heisenberg picture, the time dependence is included in the operators. Instead of the Schrödinger equation for the state vectors, we write the Heisenberg equation of motion for the spin operators

dSˆ i = i[H,ˆ Sˆ ] . (5.2.1) ~ dt i Using the Heisenberg Hamiltonian Eq. 3.4.6 and the commutation relations for the spin operators, it is straightforward to show   ˆ dSi X = − J Sˆ + gµ B × Sˆ . (5.2.2) ~ dt  ij j B 0 i j This equation is exact. We can compare it however with the classical equation of motion for the angular momentum given in Eq. 1.5.2, and we see that, by recovering the units of the spin and absorbing the Planck constant into the definition of the exchange constant, Eq. 5.2.2 can be directly translated into a classical equation of motion by taking the expectation values of the spin operators, and in particular 1 X B = B + h J Sˆ i eff 0 gµ ij j B j is the effective magnetic field including the Weiss molecular fields. We turn now to a different approximation, in which we retain for now the operator character of Sî. We are, as in the previous section, interested in the low energy excitations of the system on top of the ground state. Since the ground ˆz state is fully polarized, we expect the projection Si to remain almost constant and close to S. From Eq. 5.1.12 we see this is valid as long as NS 1, that is, the total number of spins is much larger than the number of excitations in the system. From Eq. 5.2.2 we obtain

dSˆx X i ≈ −S J Sˆy − Sˆy + gµ B Sˆy (5.2.3) ~ dt ij j i B 0 i j y dSˆ X i ≈ −S J Sˆx − Sˆx − gµ B Sˆx ~ dt ij i j B 0 i j dSˆz i ≈ 0 . ~ dt CHAPTER 5. SPIN WAVES AND MAGNONS 56

These equations are decoupled for the ladder operators   dSˆ± X i = ∓i S J Sˆ± − Sˆ± + gµ B Sˆ± , (5.2.4) ~ dt  ij i j B 0 i  j where we write the equal sign in the understanding that the equation is valid in the limit established for Eqs. 5.2.3. Eqs. 5.2.4 still couple spin operators at diﬀerent sites. To decouple them, we go once more to the Fourier representation. Using Eq. 5.1.4, for the lowering operator one obtains

dSˆ− k = iS [J(k = 0) − J(k)] Sˆ− + igµ B Sˆ−,. (5.2.5) ~ dt k B 0 k Hence we see that the equation of motion for each wave vector k is decoupled from the rest, in the spirit of a normal modes decomposition. Eq. 5.2.5 is easily solved by ˆ− ˆ iω(k)t+iαk Sk = Mke (5.2.6) with the time dependence given entirely by the exponential term, and ω(k) is given by Eq. 5.1.11 (αk is an arbitrary phase). We have therefore re-derived the dispersion relation obtained in the previous section for the one-magnon state. In the semiclassical picture and considering only one excited mode, for the real-space components of the spin we obtain

x Mk S = √ cos (k · Ri + ω(k)t) i N y Mk S = √ sin (k · Ri + ω(k)t) i N z Si = S which are the components of a plane wave with frequency ω(k). ˆ− Coming back to the Heisenberg equation of motion for Sk dSˆ− k = i[H,ˆ Sˆ−] ~ dt k from Eq. 5.2.6 we can write ˆ− ˆ ˆ− ˆ ˆ− ˆ− ~ω(k)Sk |0i = [H, Sk ]|0i = HSk |0i − Sk E0|0i and therefore we recover ˆ ˆ− ˆ− HSk |0i = [E0 − ~ω(k)] Sk |0i as in the exact result. As an example, we give the dispersion relation for spins on a cubic lattice with nearest-neighbor interactions. For this case,

J(k) = 2J (cos(kxa) + cos(kya) + cos(kza)) CHAPTER 5. SPIN WAVES AND MAGNONS 57 and therefore

~ω(k) = 2JS (3 − cos(kxa) + cos(kya) + cos(kza)) + gµBB0 . For small k we obtain a quadratic dispersion

2 2 ~ω(k) ≈ JSk a + gµBB0 , (5.2.7) which is gapped as long as B0 6= 0. Note that this dispersion is diﬀerent from the usual acoustic phonon dispersion in solids, which is linear in k. The dispersion 5.2.7 is actually similar to that encountered for ﬂexural phonon modes in materials with reduced dimensionality: for example, the out-of-plane phonon modes of a graphene membrane. The k2 dispersion is a signature of the rotational symmetry of the problem. Recommended reading: For a classical treatment, see the Section "Magnons" in Chapter 12 of the book "Introduction to solid state physics" by Charles Kittel, Ref. [7].

Check points • Derive the Heisenberg equation of motion for the spins from the Heisenberg Hamiltonian • Relate the concept of magnon from the previous section, with the semiclassical picture given by the equations of motion.

5.3 Holstein-Primakoﬀ transformation

Due to their algebra (that is, their commutation relations), angular momentum operators are difficult to treat in an interacting theory. There are however transformations which write the angular momentum operators in terms of second- quantization creation and annihilation operators, either fermionic or bosonic. The idea of these transformations is to simplify the commutation rules, so that one can use well known methods of second quantization. The price to pay is that the transformations are non linear. In this section we go over one of these transformations which is widely used, the Holstein-Primakoff transformation. Within this transformation, the angular momentum operators are written as non-linear functions of bosonic creation and annihilation operators, that is, a collection of harmonic oscillators. Before writing the transformation explicitly we remind briefly the properties † of creation (aî ) and annihilation (aî) harmonic-oscillator operators. The subscript i indicates the lattice site, that is, we have a harmonic oscillator at every site on the lattice. The commutation relations for these bosonic operators are

h †i aˆi, aˆj = δij (5.3.1)

h † †i [ˆai, aˆj] = aˆi , aˆj = 0 CHAPTER 5. SPIN WAVES AND MAGNONS 58

The Hamiltonian of a single oscillator reads 1 H = ω aˆ†aˆ + , (5.3.2) osc ~ i i i 2 where ωi is the frequency of oscillator i. If the oscillators are independent, the total Hamiltonian is simply the sum over the respective Hamiltonians. Usually however the original operators are not independent, but if the Hamiltonian is quadratic in these, one can find a linear transformation which diagonalizes the Hamiltonian, so that in the new basis the Hamiltonian is a sum of harmonic oscillators Eq. 5.3.2. A state |nii with ni "particles" at site i can be constructed † from the vacuum |0ii by applying aî (below we will identify ni with the number of flipped spins at site i). In general, we have

† where nî =a î aî is the number operator at site i. Therefore |nii is an eigenstate of Hosc with energy E(ni) = ~ωi (ni + 1/2), with ni = 0, 1, 2, .... ˆz Comparing Eqs. 5.3.3 with Eq. 3.6.4 and Si ˆ± p Si |S, mii = (S ∓ mi)(S + 1 ± mi)|S, mi ± 1i ˆz Si |S, mii = mi|S, mii we see that the creation and annihilation bosonic operators act in a similar way to the spin ladder operators, whereas the number operator is diagonal in this ˆz ˆ± † basis just as Si . However we cannot replace simply Si by aî , aî since this would not satisfy the commutation relations for the spin. This is accounted for by using a non-linear transformation, the Holstein-Primakoff transformation s √ aˆ†aˆ Sˆ+ = 2S 1 − i i aˆ i 2S i s √ aˆ†aˆ Sˆ− = 2Saˆ† 1 − i i i i 2S ˆz † Si = S − aî aî . (5.3.4)

As we anticipated, in this case the number operator counts the number of ﬂipped spins, which we can see from the last equality in Eqs. 5.3.4 ˆz Si |nii = (S − nˆi) |nii = (S − ni) |nii ≡ mi|nii .

For ni = 0, the spin is fully polarized and mi = S, hence the Holstein-Primakoﬀ vacuum |ni = 0i corresponds to the fully polarized spin state. We see however CHAPTER 5. SPIN WAVES AND MAGNONS 59

that the spectrum of nî is constrained, due to the square root in Eqs. 5.3.4 we must impose ni = 0, 1, ...2S. The maximum value of ni corresponds to the spin "fully flipped", mi = −S. We observe that r r √ nˆ √ 2S Sˆ−|m = −Si = 2Saˆ† 1 − i |n = 2Si = 2Saˆ† 1 − |n = 2Si = 0 i i i 2S i i 2S i r √ nˆ Sˆ+|m = −S − 1i = 2S 1 − i aˆ |n = 2S + 1i i i 2S i i r √ √ nˆ = 2S 2S + 1 1 − i |n = 2Si = 0 2S i and hence the ladder operators do not connect the physical subspace ni = 0, 1, ...2S with the unphysical one ni > 2S. Let’s now define r √ nˆ φ(ˆn ) = 2S 1 − i (5.3.5) i 2S and write the Heisenberg Hamiltonian Eq. 3.4.1 in terms of the Holstein- Primakoff operators, Eqs. 5.3.4. We find

2 NS J0 X X 1 X Hˆ = − + SJ nˆ − S J φ(ˆn )â aˆ†φ(ˆn ) − J nˆ nˆ (5.3.6) 2 0 i ij i i j j 2 ij i j i ij ij P with J0 = i Jij. We see that, due to Eq. 5.3.5, this Hamiltonian is not † quadratic in the aî , aî operators and therefore we cannot write it as a sum of independent harmonic oscillators. We have hence transformed our original Heisenberg Hamiltonian of interacting spins into a Hamiltonian of interacting bosons.

Check points • Write the Holstein-Primakoﬀ transformation

• What is special about it? • What is a magnon in this language? • What is the meaning of the Heisenberg Hamiltonian in terms of the bosonic operators?

5.4 Spin-wave approximation

We will now proceed to reformulate our Hamiltonian into a non-interacting term (namely, non-interacting magnons), plus interaction terms. If we expand Eq. 5.3.5 as a series nˆ nˆ2 φ(ˆn ) = 1 − i − i − ... (5.4.1) i 4S 32S2 CHAPTER 5. SPIN WAVES AND MAGNONS 60 we can write Eq. 5.3.6 also as a series

2 ∞ NS J0 X Hˆ = − + : Hˆ : (5.4.2) 2 2n n=1 with : Hˆ2n : containing n creation and n annihilation operators in normal order † † † (all aî to the left, all aî to the right, e.g. for n = 2, aî aˆjaˆmaˆl). The terms with n > 1, that is, beyond quadratic, give rise to magnon-magnon interactions as we will see below. We will however first study the spin-wave approximation, where only the quadratic, non-interacting terms are kept in the expansion Eq. 5.4.2 ˆ X X † : H2n := SJ0 nî − S Jijaî aˆj . (5.4.3) i ij This truncation of the series is justified at low temperatures, where the number of excitations (total number of flipped spins) is small compared with the total number of spins NS. For that to hold, the average number of flipped spins per site has to be small ni S and therefore√ we can approximate the square root in Eq. 5.3.5 to 1 and hence φ(ˆni) ≈ 2S. Within this approximation, the spin ladder operators are indeed approximated by simple harmonic oscillators, while the z component is kept at saturation √ + Sˆ ≈ 2Saî (5.4.4) i √ ˆ− † Si ≈ 2Saî ˆz Si ≈ S, and it can be directly seen that the Heisenberg Hamiltonian Eq. 3.4.1 is † quadratic in the aî , aî operators. This approximation is completely analogous to the one we performed when working with the equation of motion, Eq. 5.2.3. Note that, although the Hamiltonian is quadratic, it is not diagonal in i, j (see Eq. 5.4.3). Just as we did for the equations of motion, we need to go to Fourier space to obtain a diagonal Hamiltonian and therefore decoupled harmonic oscillators. In this case, we transform simply the bosonic operators 1 X −ik·Ri aˆk = √ e aî N Ri 1 X aˆ† = √ eik·Ri aˆ† k N i Ri and, within this approximation, we can show that the Heisenberg Hamiltonian reduces to ˆ X † Hsw = E(B0) + ~ω(k)âkaˆk (5.4.5) k where we have added an external magnetic field for completeness, and E(B0) and ω(k) are given by Eqs. 5.1.10 and 5.1.11 respectively. CHAPTER 5. SPIN WAVES AND MAGNONS 61

The Hamiltonian 5.4.5 describes a system of uncoupled harmonic oscillators. Its eigenstates are simply products of one-magnon states, that can be obtained † from the vacuum by applying repeatedly aˆk

n Y † k |ψSWi = aˆk |0i (5.4.6) k where nk is the number of magnons with wavevector k, eigenvalue of the number † operator in Fourier representation nˆk =a ˆkaˆk. We see that we can write the † one-magnon state deﬁned in Eq. 5.1.3 as |ki =a ˆk|0i. This state is an eigenstate of both the full Heisenberg Hamiltonian and of the non-interacting Hamiltonian 5.4.5. States with nk > 1 are however only eigenstates of 5.4.5. Recommended reading: For a good summary on second quantization, chapter 1 of the book "Fundamentals of Many-body Physics" by Wolfgang Nolting, Ref. [10].

Check points • What is the meaning of the spin-wave approximation in terms of the Holstein-Primakoﬀ transformation?

5.5 Magnon-magnon interactions

We now proceed to investigate the higher order terms (n > 1) in Eq. 5.4.2. We consider for simplicity a Heisenberg Hamiltonian with nearest neighbors interactions ˆ+ ˆ− ˆ− ˆ+ ! J X Si Sj + Si Sj Hˆ = − + SˆzSˆz . (5.5.1) n.n. 2 2 i j hiji Inserting Eqs. 5.3.4 generally we obtain  s s s s  † † † † J X aˆ aˆi aˆjaˆj aˆ aˆi aˆjaˆj Hˆ = − S 1 − i aˆ aˆ† 1 − + Saˆ† 1 − i 1 − aˆ n.n. 2  2S i j 2S i 2S 2S j hiji J X − S − aˆ†aˆ S − aˆ†aˆ . 2 i i j j hiji CHAPTER 5. SPIN WAVES AND MAGNONS 62

We now keep the ﬁrst two terms in the expansion of φ(ˆni), see Eq. 5.4.1. Therefore, simply inserting into Eq. 5.5.1

† ! † ! J X aˆ aˆi aˆjaˆj Hˆ ≈ − S 1 − i aˆ aˆ† 1 − (5.5.2) n.n. 2 4S i j 4S hiji † ! † ! J X aˆ aˆi aˆjaˆj − Saˆ† 1 − i 1 − aˆ (5.5.3) 2 i 4S 2S j hiji J X − S − aˆ†aˆ S − aˆ†aˆ . (5.5.4) 2 i i j j hiji

To be consistent with the approximation we keep terms with up to four creation/annihilation operators in Eq. 5.5.2. One obtains

JS 2 X Hˆ ≈ −Nz + JS aˆ†aˆ +a ˆ†aˆ − aˆ†aˆ − aˆ†aˆ (5.5.5) n.n. 2 i i j j i j j i hiji X 1 − J aˆ†aˆ aˆ†aˆ − aˆ†aˆ†aˆ aˆ +a ˆ†aˆ†aˆ aˆ +a ˆ†aˆ†aˆ aˆ +a ˆ†aˆ†aˆ aˆ i i j j 4 i i i j i j j j j i i i j j j i hiji (5.5.6) with the following terms in the expansion being of order 1/S or higher. We already saw that the quadratic terms in Eq. 5.5.5 can be diagonalized by going to the Fourier representation of the operators aî, after which one obtains the Hamiltonian 5.4.5 — in this case with B0 = 0 and ω(k) the corresponding one for nearest neighbors interaction. Here we pay attention to the new terms, † † for simplicity we look at one of them, e.g. the term containing aˆjaî aîaî. We CHAPTER 5. SPIN WAVES AND MAGNONS 63 denote with ∆ the nearest-neighbor vector. Hence

J X J X aˆ†aˆ†aˆ aˆ = aˆ†aˆ† aˆ aˆ 4 j i i i 4 j j+∆ j+∆ j+∆ hiji Rj ,∆ J X X = aˆ† e−ik1·Rj aˆ† e−ik2·(Rj +∆)× 4N 2 k1 k2 Rj ,∆ k1,k2,k3,k4

ik3·(Rj +∆) ik4·(Rj +∆) × aˆk3 e aˆk4 e J X X = e−i(k1+k2−k3−k4)·Rj × 4N 2 Rj ,∆ k1,k2,k3,k4 × aˆ† aˆ† aˆ aˆ e−i(k2−k3−k4)·∆ k1 k2 k3 k4 J X † † = δk +k ,k +k aˆ aˆ aˆk aˆk × 4N 1 2 3 4 k1 k2 3 4 k1,k2,k3,k4 X × e−i(k2−k3−k4)·∆ ∆ J X † † X ik1·∆ = δk +k ,k +k aˆ aˆ aˆk aˆk e . 4N 1 2 3 4 k1 k2 3 4 k1,k2,k3,k4 ∆

The last sum is simply a function of k1, which can be given explicitly once the lattice is known. For example for a cubic lattice of lattice constant a

X ik1·∆ iak1x −iak1x iak1y −iak1y iak1z −iak1z γ(k1) = e = e + e + e + e + e + e ∆

= 2 [cos (ak1x) + cos (ak1y) + cos (ak1y)] .

The term aˆ† aˆ† aˆ aˆ corresponds to two magnons with momentum k and k1 k2 k3 k4 3 k4 respectively being annihilated, and two magnons with momentum k1 and k2 being created in the interaction process. The Kronecker delta δk1+k2,k3+k4 ensures conservation of momentum, k1 + k2 = k3 + k4. The other 4-magnon interaction terms in 5.5.5 can be treated analogously. One can easily see that for long wavelength magnons (i.e. k small), the scattering cross section of such processes go as (ka)4 and is therefore small. Further interaction terms (6- magnon, etc) are suppressed by factors of increasing order in 1/S . Including dipole-dipole interactions has two main eﬀects: (i) it modiﬁes the dispersion relation ω(k), which in that case depends on the angle between the wavevector k and the equilibrium direction of the saturated spins, since the dipole-dipole interaction is anisotropic. This gives rise to a spin-wave manifold. (ii) New 3-magnon momentum-conserving interaction terms, e.g. aˆ† aˆ† aˆ or k1 k2 k3 aˆ aˆ aˆ† are allowed: one magnon can split into two, and vice-versa. k1 k2 k3 Recommended reading: Section 2.7 from "Spin Waves: Theory and Appli- cations" by Daniel D. Stancil and Anil Prabhakar, Ref. [12]. CHAPTER 5. SPIN WAVES AND MAGNONS 64

Check points • Why do we get magnon-magnon interactions?

• What conservation rules do they fulﬁll and where do they come from? Chapter 6

Magneto-Optical Eﬀects

In this chapter we will explore the interaction between light and magnetism in magnetic insulators. The coupling mechanism is the Faraday effect, in which the plane of polarization of the light rotates as it goes through a magnetized material. In turn, the light exerts a very tiny effective "magnetic field" on the spins: this is called the inverse Faraday effect and it is an example of backaction. In what follows we will go back to the classical realm to obtain the coupling term. This will allows us to, by proper quantization of the classical coupling energy term, obtain a coupling Hamiltonian between magnons and the quanta of light, photons.

6.1 Electromagnetic energy and zero-loss condition

We go back now to the full Maxwell equations, in contrast to the magnetostatic approximation we used throughout Chapter 1 ∂D ∇ × H = + j (6.1.1) ∂t F ∂B ∇ × E = − (6.1.2) ∂t ∇ · D = ρ (6.1.3) ∇ · B = 0 . (6.1.4) These equations describe completely an electromagnetic system once we give the constitutive equations

Di = εijEj (6.1.5)

Bi = µijHj where we used the Einstein convention of summation over repeated indices. These constitutive equations assume an instantaneous response of the system:

65 CHAPTER 6. MAGNETO-OPTICAL EFFECTS 66 the response does not depend on time and the system has no memory. This is referred to as a dispersionless. For an isotropic system, the permittivity and permeability tensors are diagonal and proportional to the identity, εij = ε0εrδij, µij = µ0µrδij and Eqs. 6.1.5 reduce to the scalar versions, D = ε0εrE and B = µ0µrH. We will argue that we can represent the coupling of light and magnetization just by using the permittivity tensor. Our aim now is to obtain symmetry conditions on the permittivity tensor εij in the presence of a static magnetization in the material where the light propagates. For that we will use conservation of electromagnetic energy, given by a continuity equation involving the energy ﬂux density. The instantaneous electromagnetic power per unit area is given by the Poynting vector P = E × H . (6.1.6) If we consider volume V bounded by a surface S, the energy per unit time entering the volume is given by

− P · ds ˛S where ds is an area element with vector pointing outwards. This power can be stored in the volume in the form of an energy density W , or dissipated - we denominate the dissipated power Pd. We can hence write

3 ∂W 3 3 − ∇ · Pd r = d r + Pdd r ˆV ˆV ∂t ˆV where on the LHS we have used Stokes theorem. Since the volume is arbitrary, we have ∂W ∇ · P + + P = 0 ∂t d which has the form of a continuity equation. It remains to identify the terms W and Pd in terms of the electromagnetic ﬁelds. For that we look at the Maxwell equations and we see that by taking the scalar product of 6.1.1 with E, of 6.1.2 with H, and subtracting 6.1.2 from 6.1.1, we can obtain an equation for the Poynting vector P by using the vector identity ∇ · (A × C) = C · (∇ × A) − A · (∇ × C) . (6.1.7) Putting all together we obtain the Poynting theorem ∂B ∂D ∇ · (E × H) + H · + E · + E · j = 0 (6.1.8) ∂t ∂t F ∂ (µ H ) ∂ (ε D ) ∇ · (E × H) + H ij j + E ij j + E · j = 0 , (6.1.9) i ∂t i ∂t F where in the last line we have used Eqs. 6.1.5. For dispersionless, isotropic media, we obtain µ µ ∂H2 ε ε ∂E2 ∇ · (E × H) + 0 r + 0 r + E · j = 0 , 2 ∂t 2 ∂t F CHAPTER 6. MAGNETO-OPTICAL EFFECTS 67 from where we can identify

Pd = E · jF µ µ ε ε W = 0 r H2 + 0 r E2 2 2 as the dissipated power density and instantaneous energy density stored in the magnetic and electric ﬁelds, respectively. We are interested however in time-averaged quantities. To proceed further we consider for simplicity monochromatic ﬁelds in complex notation

E(t) = Re E(ω)e−iωt H(t) = Re H(ω)e−iωt .

It can be easily shown that the time average of the product of two oscillating ﬁelds A(t) = A0 cos(ωt), B(t) = B0 cos(ωt + φ) over one period T = 2π/ω is given simply by 1 n o hA(t)B(t)i = Re A˜B˜∗ (6.1.10) T 2 where

−iωt n −iωto A(t) = Re A0e = Re Ae˜

−iφ −iωt n −iωto B(t) = Re B0e e = Re Be˜ .

As a rule, one works with the complex ﬁelds and takes the real part at the end of the calculation. In an abuse of notation, the tilde notation is dropped. We will use Eq. 6.1.10 to obtain a time average of the Poynting theorem given in Eq. 6.1.8. For that we write Maxwell equations in frequency space, in particular

∇ × H = −iωD + jF (6.1.11) ∇ × E = iωB . (6.1.12)

To obtain the Poynting theorem in complex form, we take the complex conjugate of Eq. 6.1.11 and perform the scalar product with E, and take the scalar product of Eq. 6.1.12 with H∗. Subtracting the resulting equations and using again the vector identity 6.1.7 we obtain

∗ ∗ ∗ ∗ ∇ · (E × H ) + iω (E · D − H · B) + E · jF = 0 , form where the time average is easily obtained as

∗ ∗ ∗ ∗ Re {∇ · (E × H ) + iω (E · D − H · B) + E · jF} = 0 . (6.1.13)

∗ For lossless media, h∇ · (E × H )iT must vanish, since all power that enters a volume must leave within one cycle. Moreover, if there are no free currents, the ∗ dissipated power E · jF is also zero. Therefore, in lossless media Re {iω (E · D∗ − H∗ · B)} = 0 . (6.1.14) CHAPTER 6. MAGNETO-OPTICAL EFFECTS 68

Moreover, in the frequency domain dispersive effects are included in a simple way by frequency-dependent permittivity and permeability tensors D(ω) =ε ¯(ω) · E(ω) B(ω) =µ ¯(ω) · H(ω) where the bar indicates that ε¯(ω), µ¯(ω) are matrices. Eq. 6.1.14 then can be written as iω E∗ · ε¯† − ε¯ · E + H∗ · µ¯† − µ¯ · H = 0 (6.1.15) 2 † ∗ where † indicates complex conjugate and transpose: (εij) = εji (note that the same expression can be obtained directly from Eq. 6.1.8 by replacing the real ∗ ∗ ∗ fields using Re {z} = (z + z )/2 and noting that hzziT = hz z iT = 0). We deduce therefore that for lossless media ε¯† =ε ¯ (6.1.16) µ¯† =µ ¯ , that is, the permittivity and permeability must be Hermitian matrices. Note that if the material is isotropic and ε¯(ω), µ¯(ω) can be written as in principle complex scalars ε(ω) = ε0(ω) + iε00(ω), µ(ω) = µ0(ω) + iµ00(ω), the zero-loss condition implies that the imaginary parts ε00(ω) and µ00(ω) must vanish. To define the average electromagnetic energy density in the presence of dispersion is a little bit more subtle. We give here for completeness the corresponding expression without proof 1 ∂ (ωε¯) ∂ (ωµ¯) hW i = E∗ · · E + H∗ · · H . (6.1.17) T 4 ∂ω ∂ω If ε¯ and µ¯ are independent of frequency, this expression reduces to 1 hW i = [E∗ · ε¯· E + H∗ · µ¯ · H] (6.1.18) T 4 as expected. 1. Exercise: Prove Eq. 6.1.15 starting from 6.1.14. Recommended reading: Sections 4.1 to 4.5 from "Spin Waves: Theory and Applications" by Daniel D. Stancil and Anil Prabhakar, Ref. [12]. For a more detailed treatment, Sections 80 and 96 from "Electrodynamics of continuous media" by Landau, Lifschitz, and Pitaevskii (Volume 8 of the "Course in The- oretical Physics), Ref. [?].

Check points • Obtain Eq. 6.1.13 from Maxwell equations • What is the zero loss condition? • What does it tell us about the symmetries of the permittivity tensor? CHAPTER 6. MAGNETO-OPTICAL EFFECTS 69

6.2 Permittivity tensor and magnetization

In the following section we will use the permittivity tensor ε¯ to describe the Faraday eﬀect in a magnetized sample. For that we will use the symmetry properties of ε¯ in the presence of a permanent magnetization, which breaks time reversal invariance. If we write the permittivity tensor explicitly separating the real and imaginary parts 0 00 εij = εij + iεij , the zero-loss condition 6.1.16 tells us that real and imaginary parts are respectively symmetric and antisymmetric matrices:

0 0 εij(M) = εji(M) (6.2.1) 00 00 εij(M) = −εji(M) , where we have made explicit a possible dependence on the magnetization M. On the other hand, Onsager reciprocity relations for response functions dictate how the permittivity transforms under time reversal symmetry

0 0 εij(M) = εji(−M) (6.2.2) 00 00 εij(M) = εji(−M) , where the time reversed form of ε¯ consists in transposing the matrix and at the same time inverting the magnetization vector. We see therefore that the real and imaginary parts are also symmetric and antisymmetric in the magnetization. Putting all together we obtain

0 0 0 εij(M) = εji(M) = εji(−M) (6.2.3) 00 00 00 εij(M) = −εji(M) = εji(−M) . In linear response, the permittivity depends linearly on the magnetization. This is valid as long as the effect of the magnetization on the permittivity is small. To fulfill conditions 6.2.3 to first order in the magnetization we write

εij(M) = ε0 (εrδij − ifijkMk) (6.2.4) where we have assumed the material is isotropic, and f is a small material- dependent parameter related to the Faraday rotation as we show below. A material for which the permittivity takes this form is denominated gyrotropic. 1. Exercise: Prove that Eq. 6.2.4 fulﬁlls 6.2.3. Recommended reading: To learn more about response functions and Onsager relations, section 1.3 of the book "Quantum Theory of Magnetism" by Wolfgang Nolting and Anupuru Ramakanth (2006 version), in particular the discussion leading to Eq. 1.96 in that book.

Check points • Explain how Eq. 6.2.4 is obtained CHAPTER 6. MAGNETO-OPTICAL EFFECTS 70

6.3 Faraday eﬀect

For optical frequencies one can usually safely take the permeability of a dielectric as the vacuum permeability µ0, even for a magnetic material. This amounts to neglecting the interaction of the small oscillating magnetic field part of the optical electromagnetic field with the material. In turn, the interaction between the electric part of the optical field and the magnetization in the sample is modeled by the permittivity given in Eq. 6.2.4. The magnetization M, even if it has a time dependence, it it much slower than the optical fields and therefore it is well defined. To understand how the magnetization dependent permittivity in Eq. 6.2.4 encapsulates the Faraday effect we will first derive the Fresnel equation for the optical field, starting from the Maxwell equations 6.1.11 and 6.1.12 in the absence of free currents, jF = 0. We are interested in light propagating through a material, we therefore write the electric and magnetic fields as plane waves of frequency ω and wavevector k

E(t, r) = Ee−i(ωt−k·r) H(t, r) = He−i(ωt−k·r) .

In momentum and frequency representation, Eqs. 6.1.11 and 6.1.12 read

k × E = µ0ωH (6.3.1) k × H = −ωD . (6.3.2)

Inserting Eq. 6.3.1 into 6.3.2 , using D =ε ¯·E and the product rule a×(b×c) = b(a · c) − c(a · b) one obtains

k2 k(k · E) 2 E − 2 =ε ¯· E . (6.3.3) µ0ω k This is a form of the Fresnel equation, and it determines the dispersion relation of the electromagnetic wave (that is, ω(k)) by imposing the determinant of its coeﬃcients to be zero. In components

2 k kikjEj 2 δijEj − 2 = εijEj µ0ω k and therefore 2 k kikj det 2 δij − 2 − εij = 0 . µ0ω k The term k(k·E)/k2 gives us simply the projection of the electric field along the propagation direction k/k . Whereas in vacuum the electric field E is purely transverse, in a medium the purely transverse field is actually D, see Eq. 6.3.2. We will work now however with the particular form of the permittivity given in Eq. 6.2.4, and consider the direction of propagation k/k to coincide with the CHAPTER 6. MAGNETO-OPTICAL EFFECTS 71 magnetization axis, M = Mzˆ, in which case the E field is also transverse as one can easily verify by using the resulting form of the permittivity tensor   εr −ifM 0 ε¯ = ε0  ifM εr 0  . (6.3.4) 0 0 εr

The Fresnel equation reduces therefore to

k2 ! 2 − ε0εr −ifM µ0ω 2 = 0 k ifM 2 − ε0εr µ0ω with solutions ω 2 k2 = (ε ± fM) (6.3.5) ± c r √ wehere we have used that c = 1/ ε0µ0. Inserting these solutions back into Eq. 6.3.3 with ε¯ given by Eq. 6.3.4 (in this case only two dimensional, since Ez = 0) we obtain Ex = ∓iEy , 2 with ∓ corresponding to k±. We have therefore obtained two solutions for the propagating wave along zˆ, with the same amplitude and cicularly polarized in the xy plane, but with opposite polarizations for k+ and k−: n o i(k±z−ωt) E±(z, t) = E0Re (eˆx ± ieˆy) e . (6.3.6)

To derive the Faraday rotation we consider now an EM wave propagating in the medium such that at z = 0 it is linearly polarized along xˆ with amplitude E0, −iωt E(z = 0, t) = E0eˆxe . (6.3.7) The linear combination of Eqs. 6.3.6 1 E(z, t) = [E (z, t) + E (z, t)] 2 + − fulﬁlls the condition 6.3.7. We therefore have as solution for the propagating wave E(z, t) = Re {Exeˆx + Eyeˆy}

E0 E = eik+z + eik−z x 2 E0 E = i eik+z − eik−z . y 2 After the wave propagated a distance d through the material

E k − k y = tan − + d , Ex 2 CHAPTER 6. MAGNETO-OPTICAL EFFECTS 72 which indicates that the plane of polarization of the light rotated by an amount θFd, where k − k θ = − + F 2 is the Faraday rotation per unit length. Using Eq. 6.3.5 and fM εr we obtain ω θF = √ fM . (6.3.8) 2c r

Check points • What is the Faraday eﬀect?

6.4 Magneto-optical energy

In the previous sections we saw that the magnetization in a medium modifies the permittivity tensor, which acquires and antisymmetric imaginary term due to the breaking of time reversal symmetry. As light goes through the material, it experiences this effective permittivity which we saw leads to the Faraday effect. Light and magnetization in the sample are therefore coupled. To obtain this coupling, we look at the electromagnetic energy obtained from Eq. 6.1.18 by using the permittivity Eq. 6.2.4. We see that the magnetization-dependent part of the permittivity introduces a correction to the usual electromagnetic energy expression, given by i U = − fε d3rM(r) · [E∗(r) × E(r)] . (6.4.1) MO 4 0 ˆ One can easily prove that this term is real. In terms of the Faraday rotation per unit length this can be rewritten as r εr 3 M(r) ε0 ∗ UMO = θF d r · [E (r) × E(r)] . (6.4.2) ε0 ˆ M 2iω The term ε S (r) = 0 [E∗(r) × E(r)] (6.4.3) light 2iω is called the optical spin density and it is related to the helicity of light. For example, for circularly polarized light Slight points perpendicular to the plane of polarization with a direction given by the right-hand-rule. We know from the previous section that the magnetization causes the plane of polarization of the light to rotate. From Eq. 6.4.1 we se that the light itself acts as an effective magnetic field on the magnetization (compare with the usual expression for the Zeeman energy). This gives rise to the inverse Faraday effect, which takes into account the effect of the light on the magnetization dynamics of the sample. This effect is usually small, one can show that the effective light- induced magnetic field for YIG (Yttrium Iron Garnet, a magnetic insulator, CHAPTER 6. MAGNETO-OPTICAL EFFECTS 73 widely used in both technical applications and current experiments) is of the order of 10−11T per photon/µ3. For comparison, the earth’s magnetic field is of the order of 10−6T! We will see however in the following chapter that this value can be enhanced by using an optical cavity, effectively "trapping" photons. 1. Exercise: Derive Eq. 6.4.2 starting from 6.1.18.

Check points • Obtain the correction to the electromagnetic energy if a medium is magnetized Chapter 7

Modern Topics: Cavity Optomagnonics

7.1 Quantization of the electromagnetic ﬁeld

We start by quantizing a single-mode ﬁeld in a cavity formed by perfectly con- ducting walls at z = 0 and z = L. We assume further the electric ﬁeld to be polarized along x, E = Ex(z, t)eˆx. The boundary condition therefore implies

Ex(z = 0, t) = Ex(z = L, t) = 0 . (7.1.1) From Maxwell equations in vacuum and no sources, ∂B ∇ × E = − ∇ · B = 0 (7.1.2) ∂t ∂E ∇ × B = µ ε ∇ · E = 0 (7.1.3) 0 0 ∂t we obtain a wave equation for the electric ﬁeld ∂2E ∇2E = µ ε (7.1.4) 0 0 ∂t2 which in terms of E = Ex(z, t)eˆx simpliﬁes to ∂2E (z, t) 1 ∂2E (z, t) x − x = 0 . (7.1.5) ∂z2 c2 ∂t2 The solution of Eq. 7.1.5 satisfying the boundary conditions in Eq. 7.1.1 is simply s 2 2ωn Ex(z, t) = q(t) sin (knz) (7.1.6) V ε0 with ω πn n = k = . (7.1.7) c n L

74 CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 75

In Eq. 7.1.6, q(t) has units of length and V = LS is the volume of the cavity, where S is its cross-section. From the left Eq. in 7.1.3 one obtains B = By(z, t)eˆy with s 2 µ0ε0 2ωn By(z, t) = q˙(t) cos (knz) . (7.1.8) kn V ε0 Inserting Eqs. 7.1.6 and 7.1.8 into the electromagnetic energy 1 2 1 2 EEM = dV ε0E + B 2 ˆ µ0 one obtains 1 E = ω2 q2 + p2 (7.1.9) EM 2 n where we have deﬁned p =q ˙ the canonical momentum of a "particle" of unit mass. Eq. 7.1.9 is the energy of a harmonic oscillator of unit mass. We can now proceed to quantize the theory by q → qˆ p → pˆ and imposing the commutator [ˆq, pˆ] = i~. It is convenient to introduce the bosonic creation and annihilation operators 1 aˆ = √ (ωnqˆ + ipˆ) 2~ωn † 1 aˆ = √ (ωnqˆ − ipˆ) 2~ωn which satisfy a,ˆ aˆ† in terms of which the electric and magnetic ﬁeld can be expressed as s ~ωn † Eˆx(z, t) = aˆ +a ˆ sin (knz) (7.1.10) V ε0 s ~ωn † By(z, t) = aˆ − aˆ cos (knz) . (7.1.11) V ε0 The energy Eq. 7.1.9 gives rise to the usual harmonic-oscillator Hamiltonian 1 Hˆ = ω aˆ†aˆ + . EM ~ n 2 The time dependence of the ladder operators can be obtained from the Heisen- berg equation of motion and is given by aˆ(t) =a ˆ(0)e−iωt aˆ†(t) =a ˆ†(0)eiωt . CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 76

Note that an eigenstate of the number operator nˆ =a ˆ†aˆ, nˆ|ni = n|ni, is an energy eigenstate but the electric ﬁeld operator’s expectation value vanishes

hn|Eˆx|ni = 0 ˆ2 and therefore it is not well defined. The expectation value squared field hn|Ex|ni is however finite, as one can easily prove. This reflects the uncertainty in the phase of the electric field, which is conjugate to the number operator. We call the excitation with energy ~ωn a photon. We now proceed to the quantization of multimode fields in a 3D cavity. For that, it is convenient to use the Coulomb gauge

∇ · A(r, t) = 0 , (7.1.12) in which both electric and magnetic ﬁeld can be expressed in terms of the vector potential A(r, t) ∂A(r, t) E(r, t) = − (7.1.13) ∂t B(r, t) = ∇ × A(r, t) . (7.1.14)

From Maxwell’s equations, one obtains a wave equation for the vector potential 1 ∂2A ∇2A − = 0 c2 ∂t which can be solved by separating time end space variables. It is customary to split the time dependence

A(r, t) = A(r, t)+ + A(r, t)− such that

+ X −iωkt A (r, t) = ckuk(r)e k

− X ∗ ∗ iωkt A (r, t) = ckuk(r)e k where ωk ≥ 0. The mode-functions uk are solutions of

ω 2 ∇2 + k u (r) = 0 c2 k satisfying orthonormality

∗ dV u (r)u 0 (r) = δ 0 ˆ k k k,k and the appropriate boundary conditions. The index k indicates both the mode and the polarization vector. Moreover, due to the Coulomb gauge

∇ · uk(r) = 0 . (7.1.15) CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 77

Guided by our one-mode example, we quantize replacing the amplitudes in the sum over modes by creation and annihilation operators

r h i ˆ X ~ −iωkt † iωkt A(r, t) = aˆkuk(r)e +a ˆkukˆ*(r)e 2ωkε0 k from which, using Eq. 7.1.13,we obtain the electric field operator r ωk h i ˆ ˆ + ˆ − X ~ −iωkt † iωkt E(r, t) = E (r, t) + E (r, t) = i aˆkuk(r)e − aˆkukˆ*(r)e 2ε0 k (7.1.16) with the same convention for Eˆ ±(r, t) as for Aˆ ±(r, t). Using Eq. 7.1.14 we can obtain the corresponding expression for the magnetic field. The bosonic ladder † operators aˆk, aˆk as defined are dimensionless and satisfy the usual commutation relations

h † † i [ˆak, aˆk0 ] = aˆk, aˆk0 = 0

h † i aˆk, aˆk0 = δk,k0 and the corresponding Hamiltonian is that of a collection of non-interacting harmonic oscillators X 1 H = ω aˆ† aˆ + . EM ~ k k k 2 k The sum over modes has no cutoﬀ and therefore the factor 1/2 leads to a divergence. In this notes we will not worry about that, since we will always work with diﬀerences of energies, where this factor cancels out. We will therefore simply omit this term in the following. The energy eigenstates are also eigenstates of † the number operator nˆk =a ˆkaˆk

nˆk|nki = nk|nki 1 E = ω n + . nk ~ k k 2

The states |nki form an orthonormal basis of the Hilbert space and are called number or Fock states, where nk gives the number of photons in state k.A general multi-mode state can be written as X |ψi = cn1,n2,...|n1, n2, ...i . n1,n2,...

In the simplest example, one uses periodic boundary conditions in a cubic box of side length L. In this case, 1 u (r) = eˆλeik·r k L3/2 CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 78

with k = 2π/L (nx, ny, nz), ni = 0, ±1, ±2, ... and λ = 1, 2 indicating the polarization, which from Eq. 7.1.15must fulﬁll

eˆλ · k = 0 .

One can analogously use reflecting boundary conditions, where the solutions are standing waves as in the single mode we studied above. Note that in this case the normalization factor of Eˆ(r, t) and the quantization of k will be different. 1. Exercise: Derive Eq. 7.1.4 2. Exercise: Derive Eq. 7.1.9 Recommended reading: There are several good books that treat the quantization of the electromagnetic field. A very thorough one is "Photons and Atoms: Intro- duction to Quantum Electrodynamics" by Cohen-Tannoudji, Ref. [3]. Chapter 1 of "Introductory Quantum Optics" by Gerry and Knight is a nice introduction and self-contained (Ref. [5]). Also good (and complementary to the previous references) is Chapter 13 of "Elements of Quantum Optics" by Meystre and Sargent III (Ref. [9]).

Check points • Write a general expression for the quantized electric ﬁeld

7.2 Optical cavities as open quantum systems

Cavities are usually open systems, connected to an environment both because for example the mirrors are not perfect, allowing contact to a bath of photons or/and phonons, and because we want to have access to the cavity by means of an external probe. The environment (also called bath, or reservoir) is assumed to be very large and in thermal equilibrium, and it is modeled as a collection of harmonic oscillators. The simplest Hamiltonian of the cavity plus bath is written as Hˆ = HˆS + HˆR + HÎ where HˆS , HˆR, and HÎ are the cavity, reservoir, and interaction Hamiltonians respectively ˆ † HS = ~Ωâ aˆ ˆ X ˆ†ˆ HR = ~ωkbkbk k ˆ X †ˆ ∗ˆ† HI = ~ gkaˆ bk + gkbkaˆ , k where we have taken for simplicity only one mode for the cavity aˆ with frequency Ω. The interaction Hamiltonian represents an excitation in the cavity being converted into one in the reservoir and vice-versa, with coupling constant gk. CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 79

Our aim is to obtain an effective equation of motion for the cavity mode aˆ which encapsulates the effect of the bath. This procedure is denominated to integrate out the bath, and it means that, since we are not interested in the dynamics of the bath per se, we want to eliminate these degrees of freedom and just retain the ones we are interested in, in this case, the single cavity mode. We begin by writing the Heisenberg equations of motion for both the cavity mode and the reservoir modes ˙ X ˆ aˆ(t) = −iΩâ(t) − i gkbk(t) (7.2.1) k ˆ˙ ˆ ∗ bk(t) = −iωkb(t) − igkaˆ(t) . (7.2.2) We can integrate formally Eq. 7.2.2 to obtain

t 0 ˆ ˆ −iωkt ∗ 0 0 −iωk(t−t ) bk(t) = bk(0)e − igk dt aˆ(t )e , (7.2.3) ˆ0 ˆ where the ﬁrst term corresponds to the free evolution of bk and the second one is due to the interaction with the cavity. Substituting Eq. 7.2.3 into 7.2.1 we obtain t X X 2 0 aˆ˙(t) = −iΩˆa(t) − i g ˆb (0)e−iωkt − |g | dt0aˆ(t0)e−iωk(t−t ) . (7.2.4) k k k ˆ k k 0 We now transform the cavity operators to a rotating frame with frequency Ω

Aˆ(t) =a ˆ(t)eiΩt .

We see that this transformation preserves the bosonic commutation relations h i h i Aˆ(t), Aˆ(t) = Aˆ†(t), Aˆ†(t) = 0 h i Aˆ(t), Aˆ†(t) = 1 and removes the free, fast rotating term from the equation of motion:

t ˙ X X 2 0 Aˆ(t) = −i g ˆb (0)e−i(Ω−ωk)t − |g | dt0Aˆ(t0)e−i(Ω−ωk)(t−t ) . (7.2.5) k k k ˆ k k 0 The ﬁrst term in Eq. 7.2.5 is denominated the noise operator

X ˆ −i(Ω−ωk)t Fˆ(t) = −i gkbk(0)e . k We see that this operator is composed of many diﬀerent frequencies and therefore oscillates rapidly in time. Its eﬀect on the cavity mode is that of exerting random "quantum kicks". Its expectation value for a reservoir in thermal equilibrium is easily shown to be zero hFˆ(t)iR = 0 CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 80 and therefore this operator is the quantum analog to the noise due to the environment responsible for the Brownian motion of a classical particle. The second term in Eq. 7.2.5

t X 2 0 Bˆ = − |g | dt0Aˆ(t0)e−i(Ω−ωk)(t−t ) (7.2.6) ba k ˆ k 0 is due to backaction: changes in the cavity mode aﬀect slightly the bath, which in turn acts back onto the cavity. We will see in the following that this term leads to decay of the cavity mode, which corresponds to dissipation of energy in the cavity into the environment. To analyze the second term in Eq. 7.2.5 we use that the environment volume is large, which allows us to take the continuum limit for the bath modes. We write the sum over modes directly in terms of a density of states (DOS) which we do not specify, this will depend on the details of the bath. The DOS D(ωk) gives the number of modes with frequency between ωk and ωk + dωk. In terms of the DOS, the second term in Eq. 7.2.5 is

∞ t 0 2 0 0 −i(Ω−ωk)(t−t ) Bˆba = − dωkD(ωk) |g(ωk)| dt Aˆ(t )e . (7.2.7) ˆ0 ˆ0 To perform this integral we have to resort to approximations that rely on the physics of the system. We will work with what is known as the Weisskopf- Wigner approximation, which is at its core a Markovian approximation: the evolution of the system of interest is local in time. This implies a separation of time scales between the bath, which we assume to be the fast, and the system (in our case, the cavity mode) which is slow. The information from the system that goes into the reservoir is lost, since the bath ﬂuctuates very rapidly. The system therefore is said to have no memory. The timescale of the bath is deﬁned by the inverse bandwidth 1/W . If the rate of variation Aˆ(t) is slow compared to this timescale, we can replace Aˆ(t0) → Aˆ(t) in Eq. 7.2.7, and extend the integral from t to ∞

∞ ∞ 2 −i(Ω−ωk)τ Bˆba ≈ − dωkD(ωk) |g(ωk)| Aˆ(t) dτe ˆ0 ˆ0 where we deﬁned τ = t − t0. We use now that ∞ 1 −i(Ω−ωk)τ dτe = πδ (ωk − Ω) − iP , ˆ0 ωk − Ω where the last term indicates the principal part. We neglect this term for the moment, since it leads to a frequency shift (note that its contribution is proportional to the cavity operator, and has an i in front). Evaluating the Delta function we obtain

2 Bˆba ≈ −Aˆ(t)πD(Ω) |g(Ω)| CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 81 and therefore the effective equation of motion for the cavity mode in the rotating frame is ˙ γ Aˆ(t) = − Aˆ(t) + Fˆ(t) , (7.2.8) 2 with γ = πD(Ω) |g(Ω)|2 the cavity decay rate. Eq. 7.2.8 is a quantum Langevin equation and the decay rate γ and the noise operator Fˆ(t) can be shown to fulfill the fluctuation- dissipation theorem ∞ 1 † γ = dτhFˆ (τ)Fˆ(0)iR n¯ ˆ−∞ in equilibrium, with ˆ† ˆ 1 n¯ = hb (Ω)b(Ω)iR = e~βΩ − 1 the thermal occupation of the bath at the cavity frequency. In the Markov approximation the noise correlators fulfill

† 0 00 0 00 hFˆ (t )Fˆ(t )iR = γnδ¯ (t − t ) 0 † 00 0 00 hFˆ(t )Fˆ (t )iR = γ (¯n + 1) δ (t − t ) .

In particular for the vacuum one obtains

0 † 00 0 00 h0|Fˆ(t )Fˆ (t )|0iR = γδ (t − t ) .

This delta-correlated noise shows clearly that the dynamics of the bath is fast compared to that of the system of interest, and that it has no memory since every "quantum kick" is uncorrelated with the previous one. Usually the noise operator is normalized to an operator Aˆin such that

ˆ 0 ˆ† 00 0 00 h0|Ain(t )Ain(t )|0iR = δ (t − t ) and the equation of motion is written as ˙ γ √ Aˆ(t) = − Aˆ(t) + γAˆ (t) 2 in or, in the original frame, γ √ aˆ˙(t) = −iΩˆa(t) − − aˆ(t) + γaˆ (t) . 2 in For the decay rate sometimes κ = γ/2 is used. Recommended reading: Section 15.3 from "Elements of Quantum Optics" by Meystre and Sargent III, Ref. [9]. CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 82

Check points • Write the total Hamiltonian of a cavity coupled to an environment, and explain each term • How does one obtain an eﬀective equation of motion for the cavity mode? What approximations are involved? • Write the eﬀective equation of motion for the cavity mode

• What is the meaning of κ (or γ)?

7.3 The optomagnonic Hamiltonian

We will now put together all the elements from the previous sections to derive the optomagnonic Hamiltonian, that is, the Hamiltonian for a system in which optical photons couple to magnons. For that, we will quantize the interaction term given by the Faraday effect, Eq. 6.4.1. This section and the next follow the recent work: S.Viola Kusminskiy, H. X. Tang, and F. Marquardt, Ref. [13]. ˆ+ P We can quantize the electric field following Sec. 7.1, E (r, t) = β Eβ(r)âβ(t) ˆ− P ∗ † th and correspondingly E (r, t) = β Eβ(r)âβ(t), where Eβ(r) indicates the β eigenmode of the electric field (eigenmodes are indicated with greek letters in what follows). The magnetization requires more careful consideration, since M(r) depends on the local spin operator which, in general, cannot be written as a linear combination of bosonic modes. There are however two simple cases: (i) the spin-wave approximation, which is valid for small deviations of the spins from equilibrium and, as we saw in Sec. 5.4, the Holstein-Primakoff representation can be truncated to linear order in the bosonic magnon operators, and (ii) considering the homogeneous Kittel mode M(r) = M, for which we can work simply with the resulting macrospin S. In the following we treat this second case. Although it is valid only for the homogeneous magnon mode, it allows us to capture the nonlinear dynamics of the spin. From Eq. 6.4.1 we obtain the coupling Hamiltonian

ˆ X ˆ j † HMO = ~ SjGβγ aˆβaˆγ (7.3.1) jβγ with coupling constants

j ε0f Ms ∗ Gβγ = −i jmn drEβm(r)Eγn(r) , (7.3.2) 4~S ˆ where we replaced Mj/Ms = Sˆj/S, with S the extensive total spin (scaling like the magnetic mode volume). Gj are hermitian matrices which in general cannot be simultaneously diagonalized. For simplicity, in the following we treat j the case of a strictly diagonal coupling to some optical eigenmodes (Gββ 6= 0 j but Gαβ = 0). CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 83

Consider circular polarization (R/L) in the y − z-plane. In this case, the optical spin density is perpendicular to this plane, and therefore Gx is diagonal y z while G = G = 0. The Hamiltonian HMO is then diagonal in the the basis of circularly polarized waves, e = √1 (e ∓ ie ). We choose moreover the R/L 2 y z magnetization axis along the ˆz axis. The rationale behind choosing the coupling direction perpendicular to the magnetization axis, is to maximize the coupling to the magnon mode, that is to the deviations of the magnetization with respect to the magnetization axis. The relevant spin operator is therefore Sˆx. In the case of plane waves, we quantize the electric field according to r ωj (†) ˆ+(−) X ~ +(−)ikj·r E (r, t) = +(−)i ej aˆj (t)e , 2ε0εV j where V is the volume of the cavity, kj the wave vector of mode j and we have identified the positive and negative frequency components of the field as E → + ∗ − Eˆ , E → Eˆ . The factor of ε0ε in the denominator ensures the normalization 3 2 3 2 ~ωj = ε0εhj| d r|E(r)| |ji − ε0εh0| d r|E(r)| |0i, which corresponds to the energy of a photon´ in state |ji above the´ vacuum |0i. For two degenerate (R/L) modes at frequency ω, using Eq. 6.3.8 we see that the frequency dependence cancels out and we obtain the simple form for the optomagnonic Hamiltonian ˆ † † HMO = ~GSx(âLaˆL − aˆRaˆR) with 1 c θ G = √F . S 4 ε In general however an overlap factor ξ ≤ 1 appears, which takes into account that there is a mismatch between the optical and magnonic mode volumes. For example, current experiments couple optical whispering gallery modes (WGM) in a YIG sphere to the Kittel mode. The Kittel mode is a bulk mode, and lives on the whole sphere, that is, the magnetic mode volume equals the volume of the sphere. The WGMs live however very close to the surface, and therefore its volume is smaller than the magnetic one, leading to and overlap factor ξ < 1, 1 c θ Gx = −Gx = G = √F ξ , (7.3.3) LL RR S 4 ε We now consider an incoming laser, which can drives only one of the two circular polarizations in the cavity. The total Hamiltonian of the cavity optomagnonic system is therefore given simply by † ˆ ˆ † H = −~∆â aˆ − ~ΩSz + ~GSxaˆ aˆ , (7.3.4) where aˆ† (aˆ) is the creation (annihilation) operator for the cavity mode photon which is being driven. We work in a frame rotating at the laser frequency ωlas, and ∆ = ωlas − ωcav is the detuning vs. the optical cavity frequency ωcav. In Eq. 7.3.4 we included that the dimensionless macrospin S = (Sx,Sy,Sz) has a magnetization axis along ˆz, and a Larmor precession frequency Ω which can be controlled by an external magnetic field. CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 84

1. Exercise: show that in the rotating frame the free Hamiltonian for a cavity driven mode is given by ~∆ˆa†aˆ .

Check points • Derive the optomagnonic coupling Hamiltonian

7.4 Coupled equations of motion and fast cavity limit

The coupled Heisenberg equations of motion are obtained by using a,ˆ aˆ† = 1, h i Sˆi, Sˆj = iijkSˆk. We next focus on the classical limit, where we replace the operators by their expectation values: κ a˙ = −i (GS − ∆) a − (a − α ) x 2 max η S˙ = (Ga∗a e − Ω e ) × S + G (S˙ × S) . (7.4.1) x z S

Here we introduced the cavity decay rate κ and the driving laser amplitude αmax for the optical mode. We also added an intrinsic damping for the spin ηG, which can be due to due to phonons and defects and it is material dependent. This coefficient is denominated Gilbert damping and, whereas it does not change the magnitude of the spin vector, it causes a decay of the Larmor precession to the stable equilibrium of the spin. The equation of motion for the spin without coupling to the light reduces to η S˙ = −Ω e × S + G (S˙ × S) , z S which is known as the Landau-Lifschitz-Gilbert equation. We see hence that the light acts as a kind of effective magnetic field on the spin. Actually, since the field a depends on time, and the spin-light dynamics is coupled, retardation effects cause also dissipation for the spin, in a similar way in which an environment causes the dissipation term κ for the optical field. In the following we will obtain the effective equation of motion for the spin induced by the light, "integrating out" the light field. For that we have to resort to an approximation, which is denominated the fast cavity limit, where the dynamics of the light is much faster than that of the spin (sometimes it is also called the bad cavity limit, since it implies that κ is large). That means that the photons spend in average a very short time in the cavity, during which they "see" the spin almost as static. 2 The condition for the fast cavity limit to be valid is GS˙x κ . In that case we can expand the field a(t) in powers of S˙x. We write a(t) = a0(t)+a1(t)+..., where the subscript indicates the order in S˙x. From the equation for a(t), we find that a0 fulfills the instantaneous equilibrium condition κ 1 a0(t) = αmax κ , (7.4.2) 2 2 − i (∆ − GSx(t)) CHAPTER 7. MODERN TOPICS: CAVITY OPTOMAGNONICS 85

from which we obtain the correction a1:

1 ∂a0 ˙ a1(t) = − κ Sx . (7.4.3) 2 − i (∆ − GSx) ∂Sx

2 2 To derive the eﬀective equation of motion for the spin, we replace |a| ≈ |a0| + ∗ ∗ a1a0 + a0a1 in Eq. 7.4.1 which leads to η η S˙ = B × S + opt (S˙ e × S) + G (S˙ × S) . (7.4.4) eﬀ S x x S

2 Here Beff = −Ωez +Bopt, where Bopt(Sx) = G|a0| ex is the purely static contribution and acts as an optically induced magnetic field. The second term is due to retardation effects, and it reminiscent of Gilbert damping, albeit with spin- velocity component only along ˆxdue to the chosen geometry. Both the induced field Bopt and dissipation coefficient ηopt depend explicitly on the instantaneous value of Sx(t): G κ 2 Bopt = κ 2 2 αmax ex (7.4.5) [( 2 ) + (∆ − GSx) ] 2 (∆ − GSx) ηopt = −2GκS |Bopt| κ 2 2 2 . (7.4.6) [( 2 ) + (∆ − GSx) ] These fields are highly non-linear functions of the spin. In the weak dissipation limit (ηG, ηopt 1), we can evaluate further S˙x ≈ S˙x0 from S˙ 0 = Beff × S0 and therefore obtain S˙x ≈ ΩSy. This completely determines Bopt and Γopt = ηoptΩ in terms of (Sx,Sy), which can be used to evaluate the nonlinear dynamics of the spin. Note that the optically induced dissipation can change sign! This leads to very interesting dynamics. Two distinct solutions can be found: generation of new stable fixed points (switching) and optomagnonic limit cycles.

1. Exercise: ﬁll in the steps of the derivation above.

Check points • What is the fast cavity limit?

• How do you obtain an eﬀective equation of motion for the spin, and what is the meaning of each term? Bibliography

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